Question

The function $f\left(x\right)=e^x+x$ being differentiable and one to one, has a differentiable inverse $f^{-1}\left(x\right)$, then find $\frac{d}{dx}\left(f^{-1}\right(x\left)\right)$ at the point $f\left(lo{g}_{e}2\right)$.

• A. $\frac{1}{3}$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is A $\frac{1}{3}$

Let $y=e^x+x$
on differentiating w.r.t y,
$1=(e^x+1)\frac{dx}{dy}\Rightarrow \frac{dx}{dy}=\frac{1}{1+e^x}$
$\therefore {\left(\frac{dx}{dy}\right)}_{x=lo{g}_{e}2}=\frac{1}{1+e^{log2}}=\frac{1}{3}$
or ${\left[\frac{d}{dx}\right(f^{-1}\left(x\right)\left)\right]}_{x=log2}=\frac{1}{3}$