# How do you differentiate f(x)=e^tan(1/x^2)  using the chain rule?

Dec 23, 2015

Use substitution

#### Explanation:

$f \left(x\right) = {e}^{\tan \left(\frac{1}{x} ^ 2\right)}$
Assume $t = \frac{1}{x} ^ 2$ then $p = \tan \left(t\right)$ then
$f \left(p\right) = {e}^{p}$
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dp}} \cdot \frac{\mathrm{dp}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$
$= {e}^{p} \cdot {\sec}^{2} t \cdot \left(- \frac{2}{x} ^ 3\right)$
Now back substitute

$= {e}^{\tan \left(\frac{1}{x} ^ 2\right)} \setminus \times {\sec}^{2} \left(\frac{1}{x} ^ 2\right) \setminus \times \left(\frac{- 2}{x} ^ 3\right)$

Another approach

Take $\ln$ on both sides
$\ln \left(f \left(x\right)\right) = \tan \left(\frac{1}{x} ^ 2\right)$

Now differentiate both sides

$\frac{1}{f} \setminus \times \frac{\mathrm{df}}{\mathrm{dx}} = {\sec}^{2} \left(\frac{1}{x} ^ 2\right) \setminus \times \left(\frac{- 2}{x} ^ 3\right)$
$\frac{\mathrm{df}}{\mathrm{dx}} = f \setminus \times {\sec}^{2} \left(\frac{1}{x} ^ 2\right) \setminus \times \left(\frac{- 2}{x} ^ 3\right)$
$= {e}^{\tan \left(\frac{1}{x} ^ 2\right)} \setminus \times {\sec}^{2} \left(\frac{1}{x} ^ 2\right) \setminus \times \left(\frac{- 2}{x} ^ 3\right)$