# How do you find the derivative of e^ [2 tan(sqrt x)]?

Aug 11, 2016

You do it one step at a time, keeping in mind the chain rule.

Any time you take the derivative of a function that contains a nested function (otherwise known as a composite function), take the derivative of the nested function as well.

That is, $\frac{d}{\mathrm{dx}} \left[f \left(u \left(x\right)\right)\right] = \frac{\mathrm{df} \left(u\right)}{\mathrm{du} \left(x\right)} \cdot \frac{\mathrm{du} \left(x\right)}{\mathrm{dx}}$. So:

• The derivative of $f \left(u\right) = {e}^{u}$ is ${e}^{u} \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$.
• The derivative of $f \left(u\right) = \tan u$ is ${\sec}^{2} u \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$
• The derivative of $f \left(x\right) = \sqrt{x}$ is $\frac{1}{2 \sqrt{x}}$.

Therefore:

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[{e}^{2 \tan \sqrt{x}}\right]}$

$= {e}^{2 \tan \sqrt{x}} \cdot \stackrel{\text{Chain Rule}}{\overbrace{2 \frac{d}{\mathrm{dx}} \left[\tan \sqrt{x}\right]}}$

Here, $u = \tan \sqrt{x}$, so $\textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left[u \left(x\right)\right] = {\sec}^{2} \sqrt{x} \cdot \frac{d}{\mathrm{dx}} \left[\sqrt{x}\right]}$:

$\implies {e}^{2 \tan \sqrt{x}} \cdot 2 \left({\sec}^{2} \sqrt{x} \cdot \stackrel{\text{Chain Rule Again}}{\overbrace{\frac{d}{\mathrm{dx}} \left[\sqrt{x}\right]}}\right)$

$= \cancel{2} {e}^{2 \tan \sqrt{x}} {\sec}^{2} \sqrt{x} \cdot \frac{1}{\cancel{2} \sqrt{x}}$

And here, $u = \sqrt{x}$, so now $\textcolor{g r e e n}{\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}}$:

$\implies \textcolor{b l u e}{\frac{{e}^{2 \tan \sqrt{x}} {\sec}^{2} \sqrt{x}}{\sqrt{x}}}$

That's as simple an answer as it gets, so don't be surprised if you get this. :-)