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

Jan 17, 2016

Use chain rule 3 times. It's:

$\frac{2}{x} \cdot {e}^{{\left(\ln 2 x\right)}^{2}}$

Explanation:

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

$= {e}^{{\left(\ln 2 x\right)}^{2}} \cdot 2 \cdot \frac{1}{2 x} \cdot \left(2 x\right) ' = {e}^{{\left(\ln 2 x\right)}^{2}} \cdot 2 \cdot \frac{1}{2 x} \cdot 2 =$

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

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

Explanation:

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

Differentiate both sides of the equation with respect to x

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