Problem of the Week

Updated at Jun 13, 2016 10:00 AM

This week's problem comes from the calculus category.

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

Let's begin!



\[\frac{d}{dx} \sin{x}+\sqrt{x}\]

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

2
Use Trigonometric Differentiation: the derivative of \(\sin{x}\) is \(\cos{x}\).
\[\cos{x}+(\frac{d}{dx} \sqrt{x})\]

3
Since \(\sqrt{x}={x}^{\frac{1}{2}}\), using the Power Rule, \(\frac{d}{dx} {x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}}\)
\[\cos{x}+\frac{1}{2\sqrt{x}}\]

Done
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