Problem of the Week

Updated at Apr 11, 2016 8:15 AM

This week's problem comes from the calculus category.

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

Let's begin!



\[\frac{d}{dx} \sqrt{x}\cos{x}\]

1
Use Product Rule to find the derivative of \(\sqrt{x}\cos{x}\). The product rule states that \((fg)'=f'g+fg'\).
\[(\frac{d}{dx} \sqrt{x})\cos{x}+\sqrt{x}(\frac{d}{dx} \cos{x})\]

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

3
Use Trigonometric Differentiation: the derivative of \(\cos{x}\) is \(-\sin{x}\).
\[\frac{\cos{x}}{2\sqrt{x}}-\sqrt{x}\sin{x}\]

Done
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