# Derivative of Inverse

Let $$g(x) = f^{-1}(x)$$, and $$f'(x) = \dfrac1{1+x^3}$$. Find $$g'(x)$$.

Note: $$p'(x)$$ denotes the derivative of $$p(x)$$.

Note by D K
2 years, 2 months ago

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$$\displaystyle f(g(x))=x$$ , $$\displaystyle f'(g(x))g'(x)=1\implies \frac{g'(x)}{1+g^3(x)}=1 \implies g'(x)=1+g^3(x)$$

- 2 years, 2 months ago

Note that $g'(f(x))=1+x^3\\\implies g'(x)=1+(g(x))^3$ To solve this de, use variable separition and partial fractions.

- 2 years, 2 months ago