mat01a1: derivatives of log functions and logarithmic … · mat01a1: derivatives of log functions...
TRANSCRIPT
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MAT01A1: Derivatives of Log Functions andLogarithmic Differentiation
Dr Craig
Week: 4 May 2020
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Reminder: the Chain Rule
If f and g are both differentiable and
F = f ◦ g is the composite function
defined by
F (x) = f (g(x)),
then F is differentiable and F ′ is given by
F ′(x) = f ′(g(x)).g′(x)
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Two other reminders
I ddx (a
x) = ax ln a
I When we use implicit differentiation, we
regard y as a function of x.
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In the slides that follow, I will sometimes use
y = lnx and sometimes y = `n x to denote
y = loge x. It is often good to use `n x when
writing by hand so you don’t confuse the
function with other similar symbols.
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Derivatives of Log Functions
We can use implicit differentiation to find
the derivative of the log function y = loga x.
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Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
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Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
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Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
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Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
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Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
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Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
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Derivatives of Log Functions
d
dx(loga x) =
1
x ln a
Notice that if a = e then we have
d
dx(lnx) =
1
x
An example (with chain rule):
Differentiate: y = log10(2 + cosx)
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Derivatives of Log Functions
d
dx(loga x) =
1
x ln a
Notice that if a = e then we have
d
dx(lnx) =
1
x
An example (with chain rule):
Differentiate: y = log10(2 + cosx)
![Page 14: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/14.jpg)
Derivatives of Log Functions
d
dx(loga x) =
1
x ln a
Notice that if a = e then we have
d
dx(lnx) =
1
x
An example (with chain rule):
Differentiate: y = log10(2 + cosx)
![Page 15: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/15.jpg)
Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
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Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
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Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
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Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
![Page 19: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/19.jpg)
Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
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Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
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Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
![Page 22: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/22.jpg)
Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
![Page 23: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/23.jpg)
More examples:
Differentiate the following:
1. y = ln(sinx)
2. f (x) =√lnx
3. g(x) = ln
(x + 1√x− 2
)
Solutions:
1.d
dx(ln(sinx)) =
1
sinx· ddx
(sinx) = cotx
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More examples:
Differentiate the following:
1. y = ln(sinx)
2. f (x) =√lnx
3. g(x) = ln
(x + 1√x− 2
)Solutions:
1.d
dx(ln(sinx)) =
1
sinx· ddx
(sinx) = cotx
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More examples:
Differentiate the following:
1. y = ln(sinx)
2. f (x) =√lnx
3. g(x) = ln
(x + 1√x− 2
)Solutions:
1.d
dx(ln(sinx)) =
1
sinx· ddx
(sinx) = cotx
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Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
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Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)
=d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
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Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)
=d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
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Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
![Page 30: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/30.jpg)
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)
=x− 5
2(x + 1)(x− 2)
![Page 31: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/31.jpg)
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
![Page 32: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/32.jpg)
A graph of the last example:
d
dx
(ln
x+ 1√x− 2
)=
x− 5
2(x+ 1)(x− 2)(x > 2)
![Page 33: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/33.jpg)
Differentiation so far:
I Product rule: (f.g)′ = f ′g + g′f
I Quotient rule:
(f
g
)′=
f ′g − g′f
g2
I Trig derivatives (from special limits)
I Chain Rule: F = f ◦ g,
F ′(x) = f ′(g(x)).g′(x)
I Implicit differentiation: treat y as a
function of x
I Inverse Trig derivatives
I Derivatives of Log functions
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The function f (x) = `n(x) is only defined
for x > 0. Hence its derivative f ′(x) =1
xis
only defined for x > 0. What about the
function f (x) = `n|x|?
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f (x) = `n|x| f ′(x) =1
x(both defined for x ∈ (−∞, 0) ∪ (0,∞))
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What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
![Page 37: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/37.jpg)
What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
![Page 38: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/38.jpg)
What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
![Page 39: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/39.jpg)
What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
![Page 40: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/40.jpg)
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule? Or the rule for exponential
functions?
Neither!
![Page 41: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/41.jpg)
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this?
Do we use the
power rule? Or the rule for exponential
functions?
Neither!
![Page 42: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/42.jpg)
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule?
Or the rule for exponential
functions?
Neither!
![Page 43: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/43.jpg)
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule? Or the rule for exponential
functions?
Neither!
![Page 44: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/44.jpg)
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule? Or the rule for exponential
functions?
Neither!
![Page 45: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/45.jpg)
Steps in logarithmic differentiation:
1. Take natural logarithms of both sides of
y = f (x) and use the Log Laws to
simplify the result.
2. Differentiate implicitly with respect to x.
3. Solve for y′.
![Page 46: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/46.jpg)
Steps in logarithmic differentiation:
1. Take natural logarithms of both sides of
y = f (x) and use the Log Laws to
simplify the result.
2. Differentiate implicitly with respect to x.
3. Solve for y′.
![Page 47: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/47.jpg)
Steps in logarithmic differentiation:
1. Take natural logarithms of both sides of
y = f (x) and use the Log Laws to
simplify the result.
2. Differentiate implicitly with respect to x.
3. Solve for y′.
![Page 48: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/48.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 49: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/49.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 50: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/50.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 51: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/51.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 52: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/52.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 53: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/53.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 54: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/54.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)
∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 55: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/55.jpg)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
![Page 56: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/56.jpg)
More examples:
y = (sinx)x3
and y = x√x
Try to find y′ on your own first. The
solutions are on the next two slides.
![Page 57: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/57.jpg)
Example: Differentiate y = (sinx)x3.
Solution:
ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
![Page 58: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/58.jpg)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
![Page 59: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/59.jpg)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
![Page 60: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/60.jpg)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
![Page 61: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/61.jpg)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
![Page 62: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/62.jpg)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)
∴ y′ = (sinx)x3(3x2 ln(sinx) + x3 cotx
)
![Page 63: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/63.jpg)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
![Page 64: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/64.jpg)
Example: Differentiate y = x√x.
Solution:
`n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
![Page 65: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/65.jpg)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
![Page 66: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/66.jpg)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
![Page 67: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/67.jpg)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
![Page 68: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/68.jpg)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
![Page 69: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/69.jpg)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
![Page 70: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/70.jpg)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
![Page 71: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/71.jpg)
Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way. So, instead, we
use logarithmic differentiation.
![Page 72: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/72.jpg)
Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way. So, instead, we
use logarithmic differentiation.
![Page 73: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/73.jpg)
Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way.
So, instead, we
use logarithmic differentiation.
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Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way. So, instead, we
use logarithmic differentiation.
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Logarithmic differentiation
y =x3/4√x2 + 1
(3x + 2)5
1. Take the natural logarithm of both sides
and simplify with Log Laws.
2. Differentiate both sides with respect to x
(implicitly on the LHS).
3. Solve for y′.
It is important to remember that
d
dx
(ln(g(x))
)=
g′(x)
g(x)
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Logarithmic differentiation
y =x3/4√x2 + 1
(3x + 2)5
1. Take the natural logarithm of both sides
and simplify with Log Laws.
2. Differentiate both sides with respect to x
(implicitly on the LHS).
3. Solve for y′.
It is important to remember that
d
dx
(ln(g(x))
)=
g′(x)
g(x)
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y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)= `n
(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)=
3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
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y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)
= `n(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)=
3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
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y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)= `n
(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)
=3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
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y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)= `n
(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)=
3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
![Page 81: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/81.jpg)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
![Page 82: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/82.jpg)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
![Page 83: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/83.jpg)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
![Page 84: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/84.jpg)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
![Page 85: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/85.jpg)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
![Page 86: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/86.jpg)
Logarithmic differentiation
Here are two more examples to which you
can apply logarithmic differentiation.
Attempt them on your own before looking at
the solutions:
1. f (x) =(x + 1)10
(2x− 4)8
2. h(x) =x8 cos3 x√
x− 1
![Page 87: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/87.jpg)
Solution (1.):
`n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
![Page 88: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/88.jpg)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)
∴ `n(f (x)) = `n((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
![Page 89: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/89.jpg)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)
∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
![Page 90: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/90.jpg)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
![Page 91: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/91.jpg)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
![Page 92: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/92.jpg)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
![Page 93: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/93.jpg)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
![Page 94: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/94.jpg)
The answer we have below is fine as a final
answer.
f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)If you were feeling energetic you could
simplify it further to
f ′(x) =(x + 1)9(x− 14)
128(x− 2)9
![Page 95: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/95.jpg)
Solution (2.):
`n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
![Page 96: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/96.jpg)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)
∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n(√
x− 1)
∴ `n(h(x)) = 8`n(x)+3`n(cosx)−12`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
![Page 97: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/97.jpg)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)
∴ `n(h(x)) = 8`n(x)+3`n(cosx)−12`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
![Page 98: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/98.jpg)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
![Page 99: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/99.jpg)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
![Page 100: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/100.jpg)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
![Page 101: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/101.jpg)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
![Page 102: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/102.jpg)
These last few slides are here just to give you
a better understanding of the number e.
Previously, we introduced e as follows:
e is the number such that
limh→0
eh − 1
h= 1
In other words, it is the number such that the
curve y = ex has a gradient of 1 at x = 0.
Now we want to express e as a limit.
![Page 103: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/103.jpg)
These last few slides are here just to give you
a better understanding of the number e.
Previously, we introduced e as follows:
e is the number such that
limh→0
eh − 1
h= 1
In other words, it is the number such that the
curve y = ex has a gradient of 1 at x = 0.
Now we want to express e as a limit.
![Page 104: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/104.jpg)
These last few slides are here just to give you
a better understanding of the number e.
Previously, we introduced e as follows:
e is the number such that
limh→0
eh − 1
h= 1
In other words, it is the number such that the
curve y = ex has a gradient of 1 at x = 0.
Now we want to express e as a limit.
![Page 105: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/105.jpg)
The number e as a limit
Let f (x) = `n(x). We know that f ′(x) =1
xand hence f ′(1) = 1/1 = 1.
Therefore
1 = f ′(1) = limh→0
f (1 + h)− f (1)
h
= limh→0
`n(1 + h)− `n(1)
h
= limx→0
`n(1 + x)− `n(1)
x
= limx→0
1
x`n(1 + x) = lim
x→0`n(1 + x)1/x
![Page 106: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/106.jpg)
The number e as a limit
Let f (x) = `n(x). We know that f ′(x) =1
xand hence f ′(1) = 1/1 = 1. Therefore
1 = f ′(1) = limh→0
f (1 + h)− f (1)
h
= limh→0
`n(1 + h)− `n(1)
h
= limx→0
`n(1 + x)− `n(1)
x
= limx→0
1
x`n(1 + x) = lim
x→0`n(1 + x)1/x
![Page 107: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/107.jpg)
Now,
e = e1 = elimx→0 `n(1+x)1/x
= limx→0
e`n(1+x)1/x
= limx→0
(1 + x)1/x
That is,
e = limx→0
(1 + x)1/x
As x→ 0, the bracket (1 + x) gets very close
to 1, but the 1/x will become very large.
![Page 108: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/108.jpg)
Now,
e = e1 = elimx→0 `n(1+x)1/x
= limx→0
e`n(1+x)1/x
= limx→0
(1 + x)1/x
That is,
e = limx→0
(1 + x)1/x
As x→ 0, the bracket (1 + x) gets very close
to 1, but the 1/x will become very large.
![Page 109: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/109.jpg)
Now,
e = e1 = elimx→0 `n(1+x)1/x
= limx→0
e`n(1+x)1/x
= limx→0
(1 + x)1/x
That is,
e = limx→0
(1 + x)1/x
As x→ 0, the bracket (1 + x) gets very close
to 1, but the 1/x will become very large.
![Page 110: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/110.jpg)
We have that e = limx→0
(1 + x)1/x.
Here are some calculations for x→ 0+.
![Page 111: MAT01A1: Derivatives of Log Functions and Logarithmic … · MAT01A1: Derivatives of Log Functions and Logarithmic Di erentiation Dr Craig Week: 4 May 2020](https://reader033.vdocuments.net/reader033/viewer/2022042613/5faaf6f8922b6e2a88593e6a/html5/thumbnails/111.jpg)
Prescribed tut problems:
Complete the following exercises from the
8th edition:
Ch 3.6:
2, 5, 8, 9, 13, 20, 29, 34, 39, 41, 44, 49, 52
If you are using the 7th edition:
Ch 3.6: 13 → 11