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

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

1. A

$1+{\left[g\left(x\right)\right]}^{3}$

2. B

$\frac{1}{1+{\left[g\left(x\right)\right]}^{3}}$

3. C

${\left[g\left(x\right)\right]}^{3}$

4. D

$\frac{1}{{\left[g\left(x\right)\right]}^{3}}$

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### Solution:

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

$⇒g\left(x\right)={f}^{-1}\left(x\right)⇒f\left(g\left(x\right)\right)=x$

Differentiate with respect to  $\text{'}x\text{'}$

$f\text{'}\left[g\left(x\right)\right]×g\text{'}\left(x\right)=1$

$\therefore g\text{'}\left(x\right)=\frac{1}{f\text{'}\left(g\left(x\right)\right)}=1+{\left[g\left(x\right)\right]}^{3}$

$\left(\because f\text{'}\left(x\right)=\frac{1}{1+{x}^{3}},\text{\hspace{0.17em}}f\text{'}\left(g\left(x\right)\right)=\frac{1}{1+{\left(g\left(x\right)\right)}^{3}}\right)$.

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