derivatives of inverse trig functions
DESCRIPTION
DERIVATIVES OF INVERSE TRIG FUNCTIONS. To get derivatives of inverse trigonometric functions we were able to use implicit differentiation. . - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/1.jpg)
DERIVATIVES OF INVERSE TRIG FUNCTIONS
𝑦=arc𝑠𝑖𝑛𝑥 𝑦=𝑠𝑖𝑛− 1𝑥
sin 𝑦=𝑥𝑑𝑑𝑥 sin 𝑦=
𝑑𝑑𝑥 𝑥
cos 𝑦 𝑑𝑦𝑑𝑥=1
𝑑𝑦𝑑𝑥 =
1cos 𝑦=
1√1−𝑠𝑖𝑛2 𝑦
𝑑𝑑𝑥 𝑎𝑟𝑐 sin 𝑥=
1√1−𝑥2
𝑦=arc𝑐𝑜𝑠𝑥 𝑦=𝑐𝑜𝑠− 1𝑥
cos 𝑦=𝑥𝑑𝑑𝑥 cos 𝑦=
𝑑𝑑𝑥 𝑥
−sin 𝑦 𝑑𝑦𝑑𝑥=1
𝑑𝑦𝑑𝑥 =
1−𝑠𝑖𝑛 𝑦=− 1
√1−𝑐𝑜𝑠2 𝑦𝑑𝑑𝑥 𝑎𝑟𝑐cos 𝑥=− 1
√1−𝑥2
![Page 2: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/2.jpg)
𝑦=arc 𝑡𝑎𝑛𝑥𝑦=𝑡𝑎𝑛− 1𝑥
𝑡𝑎𝑛 𝑦=𝑥𝑑𝑑𝑥 𝑡𝑎𝑛 𝑦=
𝑑𝑑𝑥 𝑥
𝑐𝑜𝑠2 𝑦+𝑠𝑖𝑛2 𝑦𝑐𝑜𝑠2𝑦
𝑑𝑦𝑑𝑥=1
𝑑𝑦𝑑𝑥 =𝑐𝑜𝑠2 𝑦=
𝑐𝑜𝑠2 𝑦𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2 𝑦
𝑑𝑑𝑥 𝑎𝑟𝑐 tan𝑥=
11+𝑥2
𝑦=arc𝑐𝑜𝑡 𝑥 𝑦=𝑐𝑜𝑡− 1𝑥
𝑐𝑜𝑡 𝑦=𝑥𝑑𝑑𝑥 co 𝑡 𝑦=
𝑑𝑑𝑥 𝑥
−𝑠𝑖𝑛2 𝑦−𝑐𝑜𝑠2 𝑦𝑠𝑖𝑛2 𝑦
𝑑𝑦𝑑𝑥=1
𝑑𝑑𝑥 𝑎𝑟𝑐cot 𝑥=− 1
1+𝑥2
𝑑𝑦𝑑𝑥 =
𝑐𝑜𝑠2 𝑦 ∙ 1𝑐𝑜𝑠2 𝑦
(𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2 𝑦 ) ∙ 1𝑐𝑜𝑠2 𝑦
𝑑𝑦𝑑𝑥 =−𝑠𝑖𝑛2 𝑦=− 𝑠𝑖𝑛2 𝑦
𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2𝑦
𝑑𝑦𝑑𝑥 =−
𝑠𝑖𝑛2 𝑦 ∙ 1𝑠𝑖𝑛2𝑦
(𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2 𝑦 ) ∙ 1𝑠𝑖𝑛2 𝑦
![Page 3: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/3.jpg)
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 ????
![Page 4: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/4.jpg)
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 ⇒ ⇒
![Page 5: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/5.jpg)
The relation
𝑔 ′ (𝑥 )= 1𝑓 ′ (𝑔 (𝑥 ))
( 𝑓 − 1)′
(𝑥 )= 1𝑓 ′ ( 𝑓 − 1 (𝑥 ))
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).
![Page 6: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/6.jpg)
A typical problem using this formula might look like this:
Given: 3 5f 3 6dfdx
Find: 1
5dfdx
1 15
6dfdx
example:
𝑓 (𝑔(𝑥))=𝑥
𝑓 ′ (𝑔(𝑥))𝑔 ′ (𝑥)=1𝑓 (3 )=5⇒𝑔 (5 )=3
.
.
![Page 7: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/7.jpg)
If , find
example:
𝑓 (𝑔(𝑥))=𝑥
𝑓 ′ (𝑔(𝑥))𝑔 ′ (𝑥)=1 .
![Page 8: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/8.jpg)
If , then
•f -1(a) = b.•(f -1)’(a) = tan .• f’(b) = tan • + = π/2
( 𝑓 − 1 )′=tan=tan (𝜋2 −𝜃)¿cot 𝜃=
1tan 𝜃=
1𝑓 ′ (𝑏)
( 𝑓 − 1 )′ (𝑎)=1
𝑓 ′ ( 𝑓 −1(𝑎)) 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑎 , 𝑠𝑜:( 𝑓 − 1 )′ (𝑥)=1
𝑓 ′ ( 𝑓 −1(𝑥))
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 .
![Page 9: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/9.jpg)
example:
If f(3) = 8, and f’(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).
![Page 10: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/10.jpg)
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. y’ = 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.
![Page 11: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/11.jpg)
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
![Page 12: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/12.jpg)
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.
![Page 13: DERIVATIVES OF INVERSE TRIG FUNCTIONS](https://reader036.vdocument.in/reader036/viewer/2022081721/56815f27550346895dcdf3c6/html5/thumbnails/13.jpg)
hafhafaf
h
)()()( lim0
Recognizing a given limit as a derivative (!!!!!!)
Example:
Example:
Example:
Tricky, isn’t it? A lot of grey cells needed.