How do you differentiate y = (x)^(e^(2x))?

Mar 18, 2018

Answer:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x} \setminus {x}^{{e}^{2 x}} \setminus \left\{\frac{1}{x} + \ln {x}^{2}\right\}$

Explanation:

We have:

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

We can take natural logarithms and use implicit logarithmic differentiation:

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

Then we can use the product rule:

$\frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x} \setminus \frac{1}{x} + 2 {e}^{2 x} \setminus \ln x$

$\therefore \frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x} \setminus \left\{\frac{1}{x} + 2 \ln x\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = y \setminus {e}^{2 x} \setminus \left\{\frac{1}{x} + \ln {x}^{2}\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{{e}^{2 x}} \setminus {e}^{2 x} \setminus \left\{\frac{1}{x} + \ln {x}^{2}\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x} \setminus {x}^{{e}^{2 x}} \setminus \left\{\frac{1}{x} + \ln {x}^{2}\right\}$