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Published byBryce Ernest McDowell Modified over 4 years ago

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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

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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

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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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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

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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).

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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.

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example: If 𝑓(𝑥)=2𝑥+cos𝑥, 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

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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.

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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

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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

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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)=3×2 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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