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 m+tan⁡m\sqrt{m}+\tan{m}?

Here are the steps:



ddmm+tan⁡m\frac{d}{dm} \sqrt{m}+\tan{m}

1
Use Sum Rule: ddxf(x)+g(x)=(ddxf(x))+(ddxg(x))\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x)).
(ddmm)+(ddmtan⁡m)(\frac{d}{dm} \sqrt{m})+(\frac{d}{dm} \tan{m})

2
Since x=x12\sqrt{x}={x}^{\frac{1}{2}}, using the Power Rule, ddxx12=12x−12\frac{d}{dx} {x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}}
12m+(ddmtan⁡m)\frac{1}{2\sqrt{m}}+(\frac{d}{dm} \tan{m})

3
Use Trigonometric Differentiation: the derivative of tan⁡x\tan{x} is sec⁡2x\sec^{2}x.
12m+sec⁡2m\frac{1}{2\sqrt{m}}+\sec^{2}m

Done
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