# Problem of the Week

## Updated at Mar 9, 2015 9:39 AM

To get more practice in calculus, we brought you this problem of the week:

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

Check out the solution below!

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

 1 Use Quotient Rule to find the derivative of $$\frac{\sqrt{x}}{\tan{x}}$$. The quotient rule states that $$(\frac{f}{g})'=\frac{f'g-fg'}{{g}^{2}}$$.$\frac{\tan{x}(\frac{d}{dx} \sqrt{x})-\sqrt{x}(\frac{d}{dx} \tan{x})}{\tan^{2}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{\frac{\tan{x}}{2\sqrt{x}}-\sqrt{x}(\frac{d}{dx} \tan{x})}{\tan^{2}x}$3 Use Trigonometric Differentiation: the derivative of $$\tan{x}$$ is $$\sec^{2}x$$.$\frac{\frac{\tan{x}}{2\sqrt{x}}-\sqrt{x}\sec^{2}x}{\tan^{2}x}$Done(tan(x)/(2*sqrt(x))-sqrt(x)*sec(x)^2)/tan(x)^2