# How do you differentiate f(x)=(ln(x-(e^(tan(x^2)))))^(3/2) using the chain rule?

Jan 3, 2016

$f ' \left(x\right) = \frac{\left(6 x {e}^{\tan} \left({x}^{2}\right) {\sec}^{2} \left({x}^{2}\right) - 3\right) \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)}}{2 {e}^{\tan} \left({x}^{2}\right) - 2 x}$

#### Explanation:

This is a mathematical bloodbath.

First Issue: the $\frac{3}{2}$ power. Deal with through chain rule and power rule: $\frac{d}{\mathrm{dx}} \left({u}^{\frac{3}{2}}\right) = \frac{3}{2} {u}^{\frac{1}{2}} = \frac{3}{2} \sqrt{u}$.

$f ' \left(x\right) = \frac{3}{2} \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)} \cdot \frac{d}{\mathrm{dx}} \left(\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)\right)$

Second Issue: the natural logarithm. Use the chain rule: $\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot u '$.

$f ' \left(x\right) = \frac{3}{2} \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)} \cdot \frac{1}{x - {e}^{\tan} \left({x}^{2}\right)} \cdot \frac{d}{\mathrm{dx}} \left(x - {e}^{\tan} \left({x}^{2}\right)\right)$

A little simplification, first...

$f ' \left(x\right) = \frac{3 \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)}}{2 \left(x - {e}^{\tan} \left({x}^{2}\right)\right)} \cdot \frac{d}{\mathrm{dx}} \left(x - {e}^{\tan} \left({x}^{2}\right)\right)$

Third Issue: the $e$ with the tangent power. Use the chain rule: $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$.

$f ' \left(x\right) = \frac{3 \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)}}{2 \left(x - {e}^{\tan} \left({x}^{2}\right)\right)} \cdot \left(1 - {e}^{\tan} \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left(\tan \left({x}^{2}\right)\right)\right)$

Fourth Issue: the tangent. Use chain rule again: $\frac{d}{\mathrm{dx}} \left(\tan \left(u\right)\right) = u ' {\sec}^{2} \left(u\right)$.

$f ' \left(x\right) = \frac{3 \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)}}{2 \left(x - {e}^{\tan} \left({x}^{2}\right)\right)} \cdot \left(1 - {e}^{\tan} \left({x}^{2}\right) {\sec}^{2} \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right)$ should be very easy to find!

$f ' \left(x\right) = \frac{3 \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)}}{2 \left(x - {e}^{\tan} \left({x}^{2}\right)\right)} \cdot \left(1 - 2 x {\sec}^{2} \left({x}^{2}\right) {e}^{\tan} \left({x}^{2}\right)\right)$

Some simplification can be done to reach...

$f ' \left(x\right) = \frac{\left(6 x {e}^{\tan} \left({x}^{2}\right) {\sec}^{2} \left({x}^{2}\right) - 3\right) \sqrt{\ln \left(x - {e}^{\tan} \left({x}^{2}\right)\right)}}{2 {e}^{\tan} \left({x}^{2}\right) - 2 x}$