Problem of the Week

Updated at Sep 22, 2014 4:39 PM

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
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