#### Need solution for RD Sharma maths class 12 chapter 11 Higher Order Derivatives exercise very short answer type question 7

\begin{aligned} &\text { The value of } \frac{d^{2} x}{d y^{2}} \text { is } \frac{-e^{x}}{\left(1+e^{x}\right)^{3}} \end{aligned}

Hint:

First differentiate given equation w.r.t to $x$

\begin{aligned} &\text { Use } \frac{d y}{d x} =\frac{1}{\frac{dy}{dx}} \end{aligned}

Given:

$y=x+e^{x}$

Explanation:

It is given that

$y=x+e^{x}$

Diff w. r .t be  $x$

\begin{aligned} &\frac{d y}{d x}=1+e^{x} \\ &\therefore \frac{d y}{d x}=\frac{1}{\frac{d y}{d x}} \\ &\therefore \frac{d x}{d y}=\frac{1}{1+e^{x}} \end{aligned}

Diff w.r. to  $y$

\begin{aligned} &\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d}{d y}\left(\frac{1}{1+e^{x}}\right) \\\\ &\frac{d^{2} x}{d y^{2}}=\frac{d}{d x}\left(\frac{1}{1+e^{x}}\right) \frac{d x}{d y} \\ &=\frac{-1}{\left(1+e^{x}\right)^{2}} e^{x}\left(\frac{1}{1+e^{x}}\right) \\ &=-\left(\frac{1}{1+e^{x}}\right)^{3} e^{x} \\ &\quad \\ &=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}} \end{aligned}

\begin{aligned} Thus,\; \; \frac{d^{2}x}{dy^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}} \end{aligned}