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Implicit Differentiation and InverseTrigonometric Functions
MATH 161 Calculus I
J. Robert Buchanan
Department of Mathematics
Summer 2019
Explicit vs. Implicit Functions
-1.0 -0.5 0.0 0.5 1.0-4
-3
-2
-1
0
x
y
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
xy
y = x2 + 2x − 3 x2 + y2 = 1
Motivation (1 of 2)
Think of y as a function y(x) then
x2 + y2 = 1x2 + (y(x))2 = 1
(y(x))2 = 1− x2
y(x) =√
1− x2 or
y(x) = −√
1− x2.
The one equation defines two implicit functions of x .
Motivation (2 of 2)
ddx
[x2 + y2
]=
ddx
[1]
ddx
[x2 + (y(x))2
]= 0
2x + 2y(x)dydx
= 0
2ydydx
= −2x
dydx
= −xy
if y 6= 0.
Implicit Differentiation
The process of differentiating both sides of an equation isknown as implicit differentiation.
When we encounter a function of y , where y is implicitly afunction of x , we use the following derivative formula (the ChainRule):
ddx
[g(y)] = g′(y)dydx
Example
Finddydx
if x ey − 3y sin x = 1.
ddx
[x ey − 3y sin x ] =ddx
[1]
ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0
(xey − 3 sin x)y ′ = 3y cos x − ey
dydx
=3y cos x − ey
xey − 3 sin x
Example
Finddydx
if x ey − 3y sin x = 1.
ddx
[x ey − 3y sin x ] =ddx
[1]
ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0
(xey − 3 sin x)y ′ = 3y cos x − ey
dydx
=3y cos x − ey
xey − 3 sin x
Example
Finddydx
if x ey − 3y sin x = 1.
ddx
[x ey − 3y sin x ] =ddx
[1]
ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0
(xey − 3 sin x)y ′ = 3y cos x − ey
dydx
=3y cos x − ey
xey − 3 sin x
Example
Finddydx
if x ey − 3y sin x = 1.
ddx
[x ey − 3y sin x ] =ddx
[1]
ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0(xey − 3 sin x)y ′ = 3y cos x − ey
dydx
=3y cos x − ey
xey − 3 sin x
ExampleFind the slope of the tangent line to the graph of
x3y2 − 4√
x = x2y at (x , y) = (4,1/2).
2 3 4 5 6
-1.0
-0.5
0.0
0.5
1.0
x
y
Solution
x3y2 − 4√
x = x2yddx
[x3y2 − 4x1/2
]=
ddx
[x2y
]
3x2y2 + 2x3y y ′ − 2√x
= 2xy + x2y ′
12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2
y ′ = − 748
Solution
x3y2 − 4√
x = x2yddx
[x3y2 − 4x1/2
]=
ddx
[x2y
]3x2y2 + 2x3y y ′ − 2√
x= 2xy + x2y ′
12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2
y ′ = − 748
Solution
x3y2 − 4√
x = x2yddx
[x3y2 − 4x1/2
]=
ddx
[x2y
]3x2y2 + 2x3y y ′ − 2√
x= 2xy + x2y ′
12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2
y ′ = − 748
Solution
x3y2 − 4√
x = x2yddx
[x3y2 − 4x1/2
]=
ddx
[x2y
]3x2y2 + 2x3y y ′ − 2√
x= 2xy + x2y ′
12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2
y ′ = − 748
ExampleFind the equation of the tangent line to the graph of
x3
y+
y3
x= 1 at (x , y) = (1/2,0.746679).
0.2 0.4 0.6 0.8 1.0
0.2
0.4
0.6
0.8
1.0
x
y
Solution (1 of 2)
x3
y+
y3
x= 1
ddx
[x3
y+
y3
x
]=
ddx
[1]
3x2y − x3y ′
y2 +3y2y ′x − y3
x2 = 0
y ′ = 0.211708
when (x , y) = (1/2,0.746679).
Solution (1 of 2)
x3
y+
y3
x= 1
ddx
[x3
y+
y3
x
]=
ddx
[1]
3x2y − x3y ′
y2 +3y2y ′x − y3
x2 = 0
y ′ = 0.211708
when (x , y) = (1/2,0.746679).
Solution (1 of 2)
x3
y+
y3
x= 1
ddx
[x3
y+
y3
x
]=
ddx
[1]
3x2y − x3y ′
y2 +3y2y ′x − y3
x2 = 0
y ′ = 0.211708
when (x , y) = (1/2,0.746679).
Solution (2 of 2)
m =y − y0
x − x0
0.211708 =y − 0.746679
x − 1/2y = 0.211708x + 0.640825
Example
Find the horizontal and vertical tangents to the graph of theequation
xy2 − 2y = 2.
First we must find y ′:
ddx
[xy2 − 2y
]=
ddx
[2]
y2 + 2xy y ′ − 2y ′ = 0y2 + 2(xy − 1)y ′ = 0
y ′ =y2
2(1− xy)
Example
Find the horizontal and vertical tangents to the graph of theequation
xy2 − 2y = 2.
First we must find y ′:
ddx
[xy2 − 2y
]=
ddx
[2]
y2 + 2xy y ′ − 2y ′ = 0y2 + 2(xy − 1)y ′ = 0
y ′ =y2
2(1− xy)
Tangents
2 = xy2 − 2y
y ′ =y2
2(1− xy)
Tangent lines are horizontal when y ′ = 0 which implies y = 0.However, when y = 0 the first equation cannot be satisfied.Thus there are no points on the curve where the tangent line ishorizontal.
Tangent lines are vertical when y ′ is undefined. This impliesxy = 1 or equivalently x = 1/y . Substituting this into the firstequation yields
2 = y − 2y = −y =⇒ (x , y) = (−1/2,−2).
Tangents
2 = xy2 − 2y
y ′ =y2
2(1− xy)
Tangent lines are horizontal when y ′ = 0 which implies y = 0.However, when y = 0 the first equation cannot be satisfied.Thus there are no points on the curve where the tangent line ishorizontal.
Tangent lines are vertical when y ′ is undefined. This impliesxy = 1 or equivalently x = 1/y . Substituting this into the firstequation yields
2 = y − 2y = −y =⇒ (x , y) = (−1/2,−2).
Illustration
2 = xy2 − 2y
-1.0 -0.5 0.0 0.5 1.0
-3.0
-2.5
-2.0
-1.5
-1.0
x
y
Example
Find y ′′ given that x3 + y3 = 1.
First we must find y ′.
ddx
[x3 + y3
]=
ddx
[1]
3x2 + 3y2y ′ = 0
y ′ = −x2
y2
Then we must differentiate a second time.
Example
Find y ′′ given that x3 + y3 = 1.
First we must find y ′.
ddx
[x3 + y3
]=
ddx
[1]
3x2 + 3y2y ′ = 0
y ′ = −x2
y2
Then we must differentiate a second time.
Example
Find y ′′ given that x3 + y3 = 1.
First we must find y ′.
ddx
[x3 + y3
]=
ddx
[1]
3x2 + 3y2y ′ = 0
y ′ = −x2
y2
Then we must differentiate a second time.
Second DerivativeSo far we have:
x3 + y3 = 1
y ′ = −x2
y2
When we differentiate again we find,
ddx[y ′]
=ddx
[−x2
y2
]
y ′′ = −2xy2 − x2(2y)y ′
(y2)2 (quotient rule)
=−2xy2 + 2x2y(−x2/y2)
y4
=−2xy2 − 2x4/y
y4
=−2xy3 − 2x4
y5
Second DerivativeSo far we have:
x3 + y3 = 1
y ′ = −x2
y2
When we differentiate again we find,
ddx[y ′]
=ddx
[−x2
y2
]y ′′ = −2xy2 − x2(2y)y ′
(y2)2 (quotient rule)
=−2xy2 + 2x2y(−x2/y2)
y4
=−2xy2 − 2x4/y
y4
=−2xy3 − 2x4
y5
Second DerivativeSo far we have:
x3 + y3 = 1
y ′ = −x2
y2
When we differentiate again we find,
ddx[y ′]
=ddx
[−x2
y2
]y ′′ = −2xy2 − x2(2y)y ′
(y2)2 (quotient rule)
=−2xy2 + 2x2y(−x2/y2)
y4
=−2xy2 − 2x4/y
y4
=−2xy3 − 2x4
y5
Right Triangle Trigonometry
Suppose θ = sin−1 x or equivalently sin θ = x , then we canpicture θ as an acute angle in a right triangle.
1x
1- x2
θ=sin-1x
Derivative of f (x) = sin−1 x
We can use the Chain Rule to find derivatives of the inversetrigonometric functions.
sin(sin−1 x) = xddx
[sin(sin−1 x)
]=
ddx
[x ]
cos(sin−1 x)ddx
[sin−1 x
]= 1√
1− x2 ddx
[sin−1 x
]= 1
ddx
[sin−1 x
]=
1√1− x2
Derivatives of the Inverse Trigonometric Functions
ddx
[sin−1 x
]=
1√1− x2
ddx
[cos−1 x
]= − 1√
1− x2
ddx
[tan−1 x
]=
11 + x2
ddx
[sec−1 x
]=
1|x |√
x2 − 1ddx
[cot−1 x
]= − 1
1 + x2
ddx
[csc−1 x
]= − 1
|x |√
x2 − 1
Examples
Find the following derivatives:
1.ddx
[cos−1 3x
]
= − 3√1− 9x2
2.ddx
[sin−1 ex
]
=ex
√1− e2x
3.ddx
[tan−1 x2
]
=2x
1 + x4
4.ddx
[sec−1 ln x
]
=1/x
| ln x |√(ln x)2 − 1
Examples
Find the following derivatives:
1.ddx
[cos−1 3x
]= − 3√
1− 9x2
2.ddx
[sin−1 ex
]
=ex
√1− e2x
3.ddx
[tan−1 x2
]
=2x
1 + x4
4.ddx
[sec−1 ln x
]
=1/x
| ln x |√(ln x)2 − 1
Examples
Find the following derivatives:
1.ddx
[cos−1 3x
]= − 3√
1− 9x2
2.ddx
[sin−1 ex
]=
ex√
1− e2x
3.ddx
[tan−1 x2
]
=2x
1 + x4
4.ddx
[sec−1 ln x
]
=1/x
| ln x |√(ln x)2 − 1
Examples
Find the following derivatives:
1.ddx
[cos−1 3x
]= − 3√
1− 9x2
2.ddx
[sin−1 ex
]=
ex√
1− e2x
3.ddx
[tan−1 x2
]=
2x1 + x4
4.ddx
[sec−1 ln x
]
=1/x
| ln x |√(ln x)2 − 1
Examples
Find the following derivatives:
1.ddx
[cos−1 3x
]= − 3√
1− 9x2
2.ddx
[sin−1 ex
]=
ex√
1− e2x
3.ddx
[tan−1 x2
]=
2x1 + x4
4.ddx
[sec−1 ln x
]=
1/x| ln x |
√(ln x)2 − 1
Homework
I Read Section 2.8I Exercises: 1–27 odd (implicit differentiation), 29–37 odd
(inverse trigonometric functions)
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