If ‘g’ is the inverse of ‘f’ and f'(x)=11+x3, then g'(x)=

If $' g'$ is the inverse of $' f'$ and $f' (x) = \frac{1}{1 + x^{3}}$ , then $g' (x) =$

A
$1 + {[g (x)]}^{3}$
B
$\frac{1}{1 + {[g (x)]}^{3}}$
C
${[g (x)]}^{3}$
D
$\frac{1}{{[g (x)]}^{3}}$

Solution:

We have $g =$ inverse of $f = f^{- 1}$

$\Rightarrow g (x) = f^{- 1} (x) \Rightarrow f (g (x)) = x$

Differentiate with respect to $' x'$

$f' [g (x)] \times g' (x) = 1$

$∴ g' (x) = \frac{1}{f' (g (x))} = 1 + {[g (x)]}^{3}$

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

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