Question

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

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

Answer 1

Answer: (B)

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