Question

If $f$ be a polynomial, then the second derivative of $f\left(e^x\right)$ is

• A. $f\text{'}\left(e^x\right)$
• B. $f\text{'}\text{'}\left(e^x\right)e^x+f\text{'}\left(e^x\right)$
• C. $f\text{'}\text{'}\left(e^x\right)e^{2x}+f\text{'}\text{'}\left(e^x\right)$
• D. $f\text{'}\text{'}\left(e^x\right)e^{2x}+f\text{'}\left(e^x\right)e^x$

Answer 1

The correct option is D

$f\text{'}\text{'}\left(e^x\right)e^{2x}+f\text{'}\left(e^x\right)e^x$

Find the second derivative of $f\left(e^x\right)$:

Let $y=f\left(e^x\right)$

Differentiate it with respect to $x:$

$\frac{dy}{dx}=f\text{'}\left(e^x\right)e^x$ [using chain rule]

Differentiate again with respect to $x:$

$\frac{d^{2}y}{d{x}^2}=f\text{'}\text{'}\left(e^x\right)e^x{e}^x+f\text{'}\left(e^x\right)e^x$ $\left[\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(uv\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{+}\mathbf{}\mathit{v}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\right]$

$=\mathbf{}\mathit{f}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}{\mathit{e}}^{\mathbf{x}}\mathbf{\right)}\mathbf{}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{}\mathbf{+}\mathbf{}\mathit{f}\mathbf{\text{'}}\mathbf{\left(}{\mathit{e}}^{\mathbf{x}}\mathbf{\right)}\mathbf{}{\mathit{e}}^{\mathbf{x}}$

Hence, Option ‘D’ is Correct.