1. 8. DERIVATIVES OF INVERSE TRIG FUNCTIONS
𝑦= arc 𝑠𝑖𝑛 𝑥
𝑦= 𝑠𝑖𝑛 −1 𝑥
𝑦= arc 𝑐𝑜𝑠 𝑥
𝑦= 𝑐𝑜𝑠 −1 𝑥
sin 𝑦 =𝑥
cos 𝑦 =𝑥
𝑑 𝑑𝑥 sin 𝑦 = 𝑑 𝑑𝑥 𝑥
𝑑 𝑑𝑥 cos 𝑦 = 𝑑 𝑑𝑥 𝑥
cos 𝑦 𝑑𝑦 𝑑𝑥 =1
− sin 𝑦 𝑑𝑦 𝑑𝑥 =1
𝑑𝑦 𝑑𝑥 = 1 cos 𝑦 = − 𝑠𝑖𝑛 2 𝑦
𝑑𝑦 𝑑𝑥 = 1 −𝑠𝑖𝑛 𝑦 =− 1 1− 𝑐𝑜𝑠 2 𝑦
𝑑 𝑑𝑥 𝑎𝑟𝑐 sin 𝑥 = 1 1− 𝑥 2
𝑑 𝑑𝑥 𝑎𝑟𝑐 cos 𝑥 =− 1 1− 𝑥 2

2. 𝑦= arc 𝑡𝑎𝑛 𝑥
𝑦= 𝑡𝑎𝑛 −1 𝑥
𝑦= arc 𝑐𝑜𝑡 𝑥
𝑦= 𝑐𝑜𝑡 −1 𝑥
𝑡𝑎𝑛 𝑦 =𝑥
𝑐𝑜𝑡 𝑦 =𝑥
𝑑 𝑑𝑥 𝑡𝑎𝑛 𝑦 = 𝑑 𝑑𝑥 𝑥
𝑑 𝑑𝑥 co𝑡 𝑦 = 𝑑 𝑑𝑥 𝑥
𝑐𝑜𝑠 2 𝑦+ 𝑠𝑖𝑛 2 𝑦 𝑐𝑜𝑠 2 𝑦 𝑑𝑦 𝑑𝑥 =1
−𝑠𝑖𝑛 2 𝑦 −𝑐𝑜𝑠 2 𝑦 𝑠𝑖𝑛 2 𝑦 𝑑𝑦 𝑑𝑥 =1
1 𝑐𝑜𝑠 2 𝑦 1 𝑐𝑜𝑠 2 𝑦
𝑑𝑦 𝑑𝑥 = 𝑐𝑜𝑠 2 𝑦 𝑠𝑖𝑛 2 𝑦+𝑐𝑜𝑠 2 𝑦
1 𝑠𝑖𝑛 2 𝑦 1 𝑠𝑖𝑛 2 𝑦
𝑑𝑦 𝑑𝑥 =− 𝑠𝑖𝑛 2 𝑦 𝑠𝑖𝑛 2 𝑦+𝑐𝑜𝑠 2 𝑦
𝑑𝑦 𝑑𝑥 = 1 𝑡𝑎𝑛 2 𝑦+1
𝑑𝑦 𝑑𝑥 =− 1 1+ 𝑐𝑜𝑡 2 𝑦
𝑑 𝑑𝑥 𝑎𝑟𝑐 tan 𝑥 = 1 1+ 𝑥 2
𝑑 𝑑𝑥 𝑎𝑟𝑐 cot 𝑥 =− 1 1+ 𝑥 2

3. 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 ????

4. DERIVATIVE OF THE INVERSE FUNCTIONS
example:
Let 𝑓 and 𝑔 be functions that are differentiable everywhere.
If 𝑔 is the inverse of 𝑓 and if 𝑔(−2) = 5 and 𝑓 ′(5) = −1/2, what is 𝑔′(−2)?
Since 𝑔 is the inverse of 𝑓 you know that
𝑓(𝑔(𝑥)) = 𝑥 holds for all 𝑥.
Differentiating both sides with respect to 𝑥, and using the the chain rule:
𝑓′ 𝑔(𝑥) 𝑔′(𝑥) = 1
𝑑𝑓 𝑑𝑔 𝑑𝑔 𝑑𝑥 =1 So
𝑓′ 𝑔 −2 𝑔′(−2) = 1
⇒ − 𝑔 ′ −2 =1
⇒ 𝑓 ′ 5 𝑔 ′ −2 =1
𝑔 ′ −2 =−2

5. But not you. 𝑓(𝑔(𝑥)) = 𝑥 The relation
𝑔 ′ 𝑥 = 1 𝑓′(𝑔 𝑥 ) 𝑓 −1 ′ 𝑥 = 1 𝑓′( 𝑓 −1 𝑥 )
used here holds whenever 𝑓 and 𝑔 are inverse functions.
Some people memorize it.
But not you.
It is easier to re-derive it any time you want to use it, by differentiating both sides of
𝑓(𝑔(𝑥)) = 𝑥
(which you should know in the middle of the night).

6. A typical problem using this formula might look like this:
example:
A typical problem using this formula might look like this:
Given:
Find:
𝑓 3 =5 ⇒ 𝑔 5 =3
𝑓′ 𝑔(𝑥) 𝑔′(𝑥) = 1
𝑓(𝑔(𝑥)) = 𝑥
𝑓′ 𝑔(5) 𝑔′(5) = 1.
𝑓′(3)𝑔′(5) = 1.

7. example:
If 𝑓(𝑥)=2𝑥+cos⁡𝑥, find ( 𝑓 −1 )’(1)
𝑓 0 =1 ⇒ 𝑔 1 =0
𝑔 ′ 1 = 1 𝑓′(0) = 1 2− sin 0
𝑔 ′ 1 = 1 2
𝑓′ 𝑔(𝑥) 𝑔′(𝑥) = 1
𝑓(𝑔(𝑥)) = 𝑥
𝑓′ 𝑔(1) 𝑔′(1) = 1
𝑓′ 𝑔(1) 𝑔′(1) = 1

8. Graphical Interpretation
If 𝑓(𝑏) = 𝑎, then
f -1(a) = b.
(f -1)’(a) = tan .
f’(b) = tan 
 +  = π/2
𝑓 −1 ′ =tan = tan 𝜋 2 −𝜃
=cot 𝜃= 1 tan 𝜃 = 1 𝑓′(𝑏)
𝑓 −1 ′ (𝑎)= 1 𝑓′ 𝑓 −1 (𝑎)
𝑓 −1 ′ (𝑥)= 1 𝑓′ 𝑓 −1 (𝑥)
𝑡𝑟𝑢𝑒 ∀ 𝑎, 𝑠𝑜:
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 𝒇 −𝟏 at 𝒙=𝒃 is the reciprocal of the slope of 𝒇 at 𝒙=𝒂.

9. example: 𝑓 𝑥 =2 𝑥 5 + 𝑥 3 +1 Find: 𝑎 𝑓 1 𝑎𝑛𝑑 𝑓′(1)
𝑎 𝑓 1 𝑎𝑛𝑑 𝑓′(1)
𝑏 𝑓 − 𝑎𝑛𝑑 𝑓 −1 ′ 4
𝑓 ′ 𝑥 =10 𝑥 4 +3 𝑥 2 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑒𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒 → 𝑓 𝑥 𝑖𝑠 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
→ 𝑓 𝑥 ℎ𝑎𝑠 𝑎𝑛 𝑖𝑛𝑣𝑒𝑟𝑠𝑒
𝑓 1 =4
𝑓 ′ 1 =13
𝑃𝑜𝑖𝑛𝑡 1,4 𝑖𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑓 𝑥 =2 𝑥 5 + 𝑥 3 +1
→𝑃𝑜𝑖𝑛𝑡 4,1 𝑖𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑓 −1 𝑥
→ 𝑓 −1 4 =1
𝑓 𝑓 −1 𝑥 = 𝑥
𝑓′ 𝑓 −1 𝑥 𝑓 −1 ′ 𝑥 =1
𝑓′ 𝑓 − 𝑓 −1 ′ 4 =1
𝑓 −1 ′ (4)= 1 𝑓′ 1 = 1 13
𝑓′ 1 𝑓 −1 ′ 4 =1

10. Since 𝑓(𝑥) is strictly increasing near 𝑥 = 8, 𝑓 ′ 𝑥 =15 𝑥 2 +1
example:
𝑓 𝑥 =5 𝑥 3 +𝑥+8
Find:
𝑓 −1 ′ 8
Since 𝑓(𝑥) is strictly increasing near 𝑥 = 8, 𝑓 ′ 𝑥 =15 𝑥 2 +1
𝑓(𝑥) has an inverse near 𝑥 =8.
𝑓 0 =8
𝑃𝑜𝑖𝑛𝑡 0,8 𝑖𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒𝑓 𝑥 =5 𝑥 3 +𝑥+8
→𝑃𝑜𝑖𝑛𝑡 8,0 𝑖𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑓 −1 𝑥
𝑓 𝑓 −1 𝑥 = 𝑥
𝑓′ 𝑓 − 𝑓 −1 ′ 8 =1
𝑓 −1 ′ (8)= 1 𝑓′ 0 =1
𝑓′ 0 𝑓 −1 ′ 8 =1

11. 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.