Question

If $y=sinx+e^x$, then $\frac{d^{2}x}{d{y}^2}$ is equal to

• A. $\frac{sinx-e^x}{{\left(cosx+e^x\right)}^2}$
• B. $\frac{sinx-e^x}{{\left(cosx+e^x\right)}^3}$
• C. $\frac{sinx-e^x}{\left(cosx+e^x\right)}$
• D. $(-sinx+e^x{)}^{-1}$

Answer 1

Answer: (B)

$y=sinx+e^x$
$\Rightarrow \frac{dy}{dx}=cosx+e^x$
$\Rightarrow \frac{dx}{dy}=\frac{1}{cosx+e^x}$
$\therefore \frac{d^{2}x}{d{y}^2}=-\frac{1}{{\left(cosx+e^x\right)}^2}\left[-sinx+e^x\right]\frac{dy}{dy}$
$=-\frac{\left(e^x-sinx\right)}{{\left(cosx+e^x\right)}^2}\times \frac{1}{cosx+e^x}$
$=\frac{sinx-e^x}{{\left(cosx+e^x\right)}^3}$