# Problem of the Week

## Updated at Sep 7, 2015 1:55 PM

This week's problem comes from the calculus category.

How would you differentiate $$\sqrt{x}\tan{x}$$?

Let's begin!

$\frac{d}{dx} \sqrt{x}\tan{x}$

 1 Use Product Rule to find the derivative of $$\sqrt{x}\tan{x}$$. The product rule states that $$(fg)'=f'g+fg'$$.$(\frac{d}{dx} \sqrt{x})\tan{x}+\sqrt{x}(\frac{d}{dx} \tan{x})$2 Since $$\sqrt{x}={x}^{\frac{1}{2}}$$, using the Power Rule, $$\frac{d}{dx} {x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}}$$$\frac{\tan{x}}{2\sqrt{x}}+\sqrt{x}(\frac{d}{dx} \tan{x})$3 Use Trigonometric Differentiation: the derivative of $$\tan{x}$$ is $$\sec^{2}x$$.$\frac{\tan{x}}{2\sqrt{x}}+\sqrt{x}\sec^{2}x$Donetan(x)/(2*sqrt(x))+sqrt(x)*sec(x)^2