# Problem of the Week

## Updated at Apr 11, 2016 8:15 AM

This week's problem comes from the calculus category.

How can we find the derivative of $$\sqrt{x}\cos{x}$$?

Let's begin!

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

 1 Use Product Rule to find the derivative of $$\sqrt{x}\cos{x}$$. The product rule states that $$(fg)'=f'g+fg'$$.$(\frac{d}{dx} \sqrt{x})\cos{x}+\sqrt{x}(\frac{d}{dx} \cos{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{\cos{x}}{2\sqrt{x}}+\sqrt{x}(\frac{d}{dx} \cos{x})$3 Use Trigonometric Differentiation: the derivative of $$\cos{x}$$ is $$-\sin{x}$$.$\frac{\cos{x}}{2\sqrt{x}}-\sqrt{x}\sin{x}$Donecos(x)/(2*sqrt(x))-sqrt(x)*sin(x)