Question

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

• A. g(x)
• B. l +g(x)
• C. $1+(g-(x{)}^3$
• D. $\frac{1}{1+{\left(g\left(x\right)\right)}^3}$
• E. 0

Answer 1

Answer: (C)

Given, $g(x)=f^{-1}(x)$
$\Rightarrow f\{g(x)\}=x$
On differentiating, w.r.t. $x$, we get
$f^{\prime }\{g(x)\}\cdot g^{\prime }(x)=1$
$\Rightarrow g^{\prime }(x)=\frac{1}{f^{\prime }\{g(x)\}}$
$\because f^{\prime }(x)=\frac{1}{1+x^3}$ (given)
$\therefore f^{\prime }\{g(x)\}=\frac{1}{1+\{g(x){\}}^3}$
Now, from Eq. (i), we get
$g^{\prime }(x)=1+\{g(x){\}}^3$