#### Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives exercise Fill in the blanks question 5

Answer:  $\frac{d^{2} y}{d x^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}$

Hint: Differentiate given equation w.r.t x. Use  $\frac{d x}{d y}=\frac{1}{d y / d x}$

Given:  $y=x+e^{x}$

Solution:

It is given that:  $y=x+e^{x}$

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

$\frac{d x}{d y}=\frac{1}{d y / d x} \\$

$\therefore \frac{d x}{d y}=\frac{1}{1+e^{x}}$                                                                 …(i)

Again, differentiating w.r.t x:

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

$\frac{d^{2} x}{d y^{2}}=\frac{d}{d x}\left(\frac{1}{1+e^{x}}\right) \frac{d x}{d y} \\$

$=\left(\frac{-1}{\left(1+e^{x}\right)^{2}}\right) \frac{d}{d x}\left(e^{x}\right) \cdot\left(\frac{1}{1+e^{x}}\right)$                           [ using (i) ]

$=\left(\frac{-1}{\left(1+e^{x}\right)^{2}}\right)\left(e^{x}\right) \cdot\left(\frac{1}{1+e^{x}}\right)$

Thus, $\frac{d^{2} x}{d y^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}$