Problem of the Week

Updated at Jul 3, 2017 11:52 AM

How would you differentiate 2x+cot⁡x2x+\cot{x}?

Below is the solution.



ddx2x+cot⁡x\frac{d}{dx} 2x+\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)).
(ddx2x)+(ddxcot⁡x)(\frac{d}{dx} 2x)+(\frac{d}{dx} \cot{x})

2
Use Power Rule: ddxxn=nxn−1\frac{d}{dx} {x}^{n}=n{x}^{n-1}.
2+(ddxcot⁡x)2+(\frac{d}{dx} \cot{x})

3
Use Trigonometric Differentiation: the derivative of cot⁡x\cot{x} is −csc⁡2x-\csc^{2}x.
2−csc⁡2x2-\csc^{2}x

Done
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