Question

Suppose $f$ is differentiable function such that $f\left(g\right(x\left)\right)=x^2$ and $f^{\prime }\left(x\right)=1+\left(f\right(x){)}^2$. Then the value of $g^{\prime }\left(2\right)$ is

• A. $\frac{1}{17}$
• B. $\frac{2}{17}$
• C. $\frac{3}{17}$
• D. $\frac{4}{17}$

Answer 1

The correct option is D $\frac{4}{17}$

$f\left(g\right(x\left)\right)=x^2$
Differentiating w.r.t $x$, we get
$f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)=2x\cdots \left(1\right)$
Also, $f^{\prime }\left(x\right)=1+\left(f\right(x){)}^2$
$\Rightarrow f^{\prime }\left(g\right(x\left)\right)=1+\left(f\right(g\left(x\right)){)}^2\cdots (2)$
Substituting $\left(2\right)$ in $\left(1\right)$, we get
$(1+(x^2{)}^2)\cdot g^{\prime }(x)=2x$
$\Rightarrow g^{\prime }\left(x\right)=\frac{2x}{1+x^4}$
$\Rightarrow g^{\prime }\left(2\right)=\frac{4}{1+16}=\frac{4}{17}$