# Given y = e^((ln x)^2) how do you find find y'(e)?

May 2, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{{\ln}^{2} x} \ln x}{x} \implies y ' \left(e\right) = 2$

#### Explanation:

We have:

$y = {e}^{{\ln}^{2} x}$

Take Natural Logarithms of both sides:

$\ln y = {\left(\ln x\right)}^{2}$

Differentiate Implicitly, and apply the chain rule:

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\ln x\right) \cdot \left(\frac{1}{x}\right)$

Which we can rearrange to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 y \ln x}{x}$
$\text{ } = \frac{2 {e}^{{\ln}^{2} x} \ln x}{x}$

So then, when $x = e$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{{\ln}^{2} e} \ln e}{e}$
$\text{ } = \frac{2 {e}^{1} \cdot 1}{e}$
$\text{ } = 2$

May 2, 2017

Recall that $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$. Then, through the chain rule, $\frac{d}{\mathrm{dx}} {e}^{f} \left(x\right) = {e}^{f} \left(x\right) f ' \left(x\right)$.

So:

$y \left(x\right) = {e}^{{\left(\ln x\right)}^{2}}$

$y ' \left(x\right) = {e}^{{\left(\ln x\right)}^{2}} \frac{d}{\mathrm{dx}} {\left(\ln x\right)}^{2}$

$\textcolor{w h i t e}{y ' \left(x\right)} = {e}^{{\left(\ln x\right)}^{2}} \left(2 \left(\ln x\right)\right) \frac{d}{\mathrm{dx}} \ln x$

$\textcolor{w h i t e}{y ' \left(x\right)} = \frac{2 {e}^{{\left(\ln x\right)}^{2}} \ln x}{x}$

Then:

$y ' \left(e\right) = \frac{2 {e}^{{\left(\ln e\right)}^{2}} \ln e}{e}$

$\textcolor{w h i t e}{y ' \left(e\right)} = \frac{2 {e}^{1} \left(1\right)}{e}$

$\textcolor{w h i t e}{y ' \left(e\right)} = 2$