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# Derivative of Inverse Functions

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If the function f is defined by f(x) = 2x+sinx and the function g is the inverse of f, then g'(2)=
A) 0.324
B) 0.342
C) 0.360
D) 0.378

Dec 13, 2019

#1
+24350
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If the function f is defined by f(x) = 2x+sinx and the function g is the inverse of f, then g'(2)=

$$\begin{array}{|rcll|} \hline \mathbf{ g\Big(f(x)\Big) } &=& \mathbf{ x } \quad | \quad \text{chain rule} \\ g'\Big(f(x)\Big)f'(x) &=& 1\\ \mathbf{ g'\Big(f(x)\Big) } &=& \mathbf{ \dfrac{1}{f'(x)} } \\\\ && f(x) = 2 = 2x+\sin(x) \Rightarrow x = 0.68403665667782943943\ldots \text{ found with Newton's Method } \\ g'(2) &=& \dfrac{1}{f'(0.68403665667782943943)} \quad | \quad f'(x) = 2+\cos(x) \\ g'(2) &=& \dfrac{1}{2+\cos(0.68403665667782943943)} \\ \mathbf{g'(2)} &=& \mathbf{0.360356702385802856117...} \\ \hline \end{array}$$

Dec 13, 2019