Question

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

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

Answer 1

The correct option is A

$e^x$

The explanation for the correct explanation:

The given equation is $y=x+e^x$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(x+e^x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{dx}{dx}+\frac{d}{dx}\left(e^x\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(u+v\right)\mathbf{=}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ \Rightarrow \frac{dy}{dx}=1+e^x \end{gathered}$$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(1+e^x\right) \\ \Rightarrow \frac{{\mathrm{d}}^{2}y}{d{x}^2}=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(e^x\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{)}\mathbf{=}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ \Rightarrow \frac{{\mathrm{d}}^{2}y}{d{x}^2}=e^x \end{gathered}$$

Therefore, $\frac{{\mathrm{d}}^{2}y}{d{x}^2}=e^x$.

Hence, option (A) is the correct opttion.