Question

# Let $$\mathrm{g}(\mathrm{x})$$ be the inverse of the function $$\mathrm{f}(\mathrm{x})$$ and $$\displaystyle \mathrm{f}'(\mathrm{x})=\frac{1}{1+\mathrm{x}^{3}}$$ Then $$\mathrm{g'}(\mathrm{x})$$ is

A
11+(g(x))3
B
11+(f(x))3
C
1+(g(x))3
D
1+(f(x))3

Solution

## The correct option is C $$1+(\mathrm{g}(\mathrm{x}))^{3}$$Given that $$g(x)$$ is inverse of $$f(x)$$.$$\displaystyle \Rightarrow g(f(x)) = x \Rightarrow g'(f(x)). f'(x)= 1$$$$g'(f(x))=\dfrac{1}{f'(x)}=1+x^3$$$$f(x)=y$$$$g'(y)=\dfrac{1}{f'(x)}=1+x^3$$$$x=f^-1(y)=g(y)$$..............................................................(i)$$g'(y)=1+g^3(y)$$ by (i)$$g'(x)=1+g^3(x)$$Mathematics

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