to get derivatives of inverse trigonometric functions we were able to use implicit differentiation....
TRANSCRIPT
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DERIVATIVES OF INVERSE TRIG FUNCTIONS
π¦=arcπ πππ₯ π¦=π ππβ 1π₯
sin π¦=π₯
πππ₯sin π¦=
πππ₯
π₯
cos π¦ππ¦ππ₯
=1
ππ¦ππ₯
=1
cosπ¦=
1
β1βπ ππ2 π¦πππ₯
πππ sin π₯=1
β1βπ₯2
π¦=arcπππ π₯ π¦=πππ β 1π₯
cos π¦=π₯πππ₯cosπ¦=
πππ₯
π₯
βsin π¦ππ¦ππ₯
=1
ππ¦ππ₯
=1
βπ ππ π¦=β
1
β1βπππ 2 π¦πππ₯
πππcosπ₯=β1
β1βπ₯2
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π¦=arc π‘πππ₯π¦=π‘ππβ 1π₯
π‘ππ π¦=π₯
πππ₯
π‘ππ π¦=πππ₯
π₯
πππ 2 π¦+π ππ2 π¦πππ 2π¦
ππ¦ππ₯
=1
ππ¦ππ₯
=πππ 2 π¦= πππ 2 π¦π ππ2 π¦+πππ 2 π¦
πππ₯
πππ tanπ₯=1
1+π₯2
π¦=arcπππ‘ π₯ π¦=πππ‘β 1π₯
πππ‘ π¦=π₯
πππ₯coπ‘ π¦=
πππ₯
π₯
βπ ππ2 π¦βπππ 2 π¦π ππ2 π¦
ππ¦ππ₯
=1
πππ₯
πππcot π₯=β1
1+π₯2
ππ¦ππ₯
=πππ 2 π¦ β
1
πππ 2 π¦
(π ππ2 π¦+πππ 2 π¦ ) β 1πππ 2 π¦
ππ¦ππ₯
=βπ ππ2 π¦=β π ππ2 π¦π ππ2 π¦+πππ 2π¦
ππ¦ππ₯
=βπ ππ2 π¦ β
1
π ππ2π¦
(π ππ2 π¦+πππ 2 π¦ ) β 1π ππ2 π¦
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To get derivatives of inverse trigonometric functions we were able to use implicit
differentiation.
Sometimes it is not possible/plausible to explicitly find inverse function, but we still want to find derivative of inverse function at certain point (slope).
QUESTION: What is the relationship between derivatives of a function and its inverse ????
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Let and be functions that are differentiable everywhere.If is the inverse of and if and , what is ?
DERIVATIVE OF THE INVERSE FUNCTIONSexample:
Since is the inverse of you know that
holds for all .
Differentiating both sides with respect to , and using the the chain rule:
.
So β β
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The relation
π β² (π₯ )= 1π β² (π (π₯ ))
( π β 1)β²
(π₯ )= 1π β² ( π β 1 (π₯ ))
used here holds whenever and are inverse functions. Some people memorize it. It is easier to re-derive it any time you want to use it, by differentiating both sides of
(which you should know in the middle of the night).
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A typical problem using this formula might look like this:
Given: 3 5f 3 6df
dx Find:
1
5df
dx
1 15
6
df
dx
example:
π (π(π₯))=π₯
π β² (π(π₯))π β² (π₯)=1
π (3 )=5βπ (5 )=3
.
.
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If , find
example:
π (π(π₯))=π₯
π β² (π(π₯))π β² (π₯)=1 .
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If , then
β’f -1(a) = b.β’(f -1)β(a) = tan .
β’ fβ(b) = tan
β’ + = Ο/2
( π β 1 )β²=tan=tan (π2 βπ)ΒΏcot π=
1tan π
=1
π β² (π)
( π β 1 )β² (π)=1
π β² ( π β1(π)) π‘ππ’π πππ πππ¦ π , π π:( πβ 1 )β² (π₯)=
1
π β² ( π β1(π₯))
Graphical Interpretation
Derivative of the inverse function at a point is the reciprocal of the derivative of the function at the corresponding point.
Slope of the line tangent to at is the reciprocal of the slope of at .
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example:
If f(3) = 8, and fβ(3)= 5, what do we know about f-1 ?
Since f passes through the point (3, 8), f-1 must pass through the point (8, 3). Furthermore, since the graph of f has slope 5 at (3, 8), the graph of f-1 must have slope 1/5 at (8, 3).
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If f (x)= 2x5 + x3 + 1, find (a) f (1) and f '(1) and (b) (f -1 )(4) and (f -1)'(4).
y=2x5 + x3 + 1. yβ = 10x4 + 3x2 is positive everywhere y is strictly increasing, thus f (x) has an inverse.
example:
f (1)=2(1)5 + (1)3 +1=4f '(x)=10x4 + 3x2
f '(1)=10(1)4 +3(1)2 =13
Since f(1)=4 implies the point (1, 4) is on the curve f(x)=2x5 +x3 +1, therefore, the point (4, 1) (which is the reflection of (1, 4) on y =x) is on the curve (f -1)(x). Thus, (f -1)(4)=1.
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example:
If f(x)=5x3 + x +8, find (f -1) '(8). Since y is strictly increasing near x =8, f(x) has an inverse near x =8.Note that f(0)=5(0)3 +0+8=8 which implies the point (0, 8) is on the curve of f(x). Thus, the point (8, 0) is on the curve of (f -1)(x).
f '(x)=15x2 +1f '(0)=1
http://www.millersville.edu/~bikenaga/calculus/inverse-functions/inverse-functions.html
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We have been more careful than usual in our statement of the differentiability result for inverse functions. You should notice that the differentiation formula for the inverse function involves division by f '(f -1(x)). We must therefore assume that this value is not equal to zero. There is also a graphical explanation for this necessity.
Example. The graphs of the cubing function f(x) = x3 and its inverse (the cube root function) are shown below.
Notice that f '(x)=3x2 and so f '(0)=0. The cubing function has a horizontal tangent line at the origin. Taking cube roots we find that f -1(0)=0 and so f '(f -1(0))=0. The differentiation formula for f -1 can not be applied to the inverse of the cubing function at 0 since we can not divide by zero. This failure shows up graphically in the fact that the graph of the cube root function has a vertical tangent line (slope undefined) at the origin.
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h
afhafaf
h
)()()( lim
0
Recognizing a given limit as a derivative (!!!!!!)
Example:
Example:
Example:
Tricky, isnβt it? A lot of grey cells needed.