Problem of the Week

Updated at Jul 29, 2019 4:42 PM

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

Below is the solution.



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

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

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

3
Use Trigonometric Differentiation: the derivative of \(\sin{x}\) is \(\cos{x}\).
\[\frac{1}{2\sqrt{x}}+\cos{x}\]

Done
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