Problem of the Week

Updated at Nov 23, 2020 2:26 PM

For this week we've brought you this calculus problem.

How can we solve for the derivative of \(\sqrt{m}+\tan{m}\)?

Here are the steps:



\[\frac{d}{dm} \sqrt{m}+\tan{m}\]

1
Use Sum Rule: \(\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))\).
\[(\frac{d}{dm} \sqrt{m})+(\frac{d}{dm} \tan{m})\]

2
Since \(\sqrt{x}={x}^{\frac{1}{2}}\), using the Power Rule, \(\frac{d}{dx} {x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}}\)
\[\frac{1}{2\sqrt{m}}+(\frac{d}{dm} \tan{m})\]

3
Use Trigonometric Differentiation: the derivative of \(\tan{x}\) is \(\sec^{2}x\).
\[\frac{1}{2\sqrt{m}}+\sec^{2}m\]

Done
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