# Problem of the Week

## Updated at Jul 29, 2019 4:42 PM

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

Below is the solution.

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

 1 Use Sum Rule: $$\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))$$.$(\frac{d}{dx} \sqrt{x})+(\frac{d}{dx} \sin{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{1}{2\sqrt{x}}+(\frac{d}{dx} \sin{x})$3 Use Trigonometric Differentiation: the derivative of $$\sin{x}$$ is $$\cos{x}$$.$\frac{1}{2\sqrt{x}}+\cos{x}$Done1/(2*sqrt(x))+cos(x)