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+x\sin{x}+\sqrt{x}?

Let's begin!



ddxsin⁡x+x\frac{d}{dx} \sin{x}+\sqrt{x}

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)).
(ddxsin⁡x)+(ddxx)(\frac{d}{dx} \sin{x})+(\frac{d}{dx} \sqrt{x})

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

3
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}}
cos⁡x+12x\cos{x}+\frac{1}{2\sqrt{x}}

Done
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