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