Problem of the Week

Updated at Jan 18, 2016 9:00 AM

This week's problem comes from the calculus category.

How can we find the derivative of sin⁡x+cot⁡x\sin{x}+\cot{x}?

Let's begin!



ddxsin⁡x+cot⁡x\frac{d}{dx} \sin{x}+\cot{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)+(ddxcot⁡x)(\frac{d}{dx} \sin{x})+(\frac{d}{dx} \cot{x})

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

3
Use Trigonometric Differentiation: the derivative of cot⁡x\cot{x} is −csc⁡2x-\csc^{2}x.
cos⁡x−csc⁡2x\cos{x}-\csc^{2}x

Done
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