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

Apr 3, 2016

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

#### Explanation:

We have the chain rule working on multiple levels. The first level is that we have an exponential function in the form ${e}^{u}$, where $u = \tan \left(\sqrt{x}\right)$.

The chain rule states that

$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$

So, we see that

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

Now, we have to deal with differentiating the tangent function in the form $\tan \left(u\right)$, where $u = \sqrt{x}$.

Again, through the chain rule, we see that

$\frac{d}{\mathrm{dx}} \left(\tan \left(u\right)\right) = {\sec}^{2} \left(u\right) \cdot u '$

This gives us

$\frac{d}{\mathrm{dx}} \left(\tan \left(\sqrt{x}\right)\right) = {\sec}^{2} \left(\sqrt{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right)$

To find $\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right)$, recall that $\sqrt{x} = {x}^{\frac{1}{2}}$, so $\frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right) = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$.

Combining everything we've found, we see that

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

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