# Problem of the Week

## Updated at Jun 13, 2016 10:00 AM

This week's problem comes from the calculus category.

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

Let's begin!

$\frac{d}{dx} \sin{x}+\sqrt{x}$

 1 Use Sum Rule: $$\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))$$.$(\frac{d}{dx} \sin{x})+(\frac{d}{dx} \sqrt{x})$2 Use Trigonometric Differentiation: the derivative of $$\sin{x}$$ is $$\cos{x}$$.$\cos{x}+(\frac{d}{dx} \sqrt{x})$3 Since $$\sqrt{x}={x}^{\frac{1}{2}}$$, using the Power Rule, $$\frac{d}{dx} {x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}}$$$\cos{x}+\frac{1}{2\sqrt{x}}$Donecos(x)+1/(2*sqrt(x))