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 = 1 1− 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

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

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 ????

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 ⇒ − 1 2 𝑔 ′ −2 =1 ⇒ 𝑓 ′ 5 𝑔 ′ −2 =1 𝑔 ′ −2 =−2

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).

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.

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

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.

example: 𝑓 𝑥 =2 𝑥 5 + 𝑥 3 +1 Find: 𝑎 𝑓 1 𝑎𝑛𝑑 𝑓′(1) 𝑎 𝑓 1 𝑎𝑛𝑑 𝑓′(1) 𝑏 𝑓 −1 4 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 −1 4 𝑓 −1 ′ 4 =1 𝑓 −1 ′ (4)= 1 𝑓′ 1 = 1 13 𝑓′ 1 𝑓 −1 ′ 4 =1

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 −1 8 𝑓 −1 ′ 8 =1 𝑓 −1 ′ (8)= 1 𝑓′ 0 =1 𝑓′ 0 𝑓 −1 ′ 8 =1 http://www.millersville.edu/~bikenaga/calculus/inverse-functions/inverse-functions.html

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.