# How do you differentiate e^((ln2x)^2+x)  using the chain rule?

Nov 2, 2016

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

#### Explanation:

It requires several application of the chain rule , which I will do implicity:

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