Problem of the Week

Updated at Jun 12, 2017 3:52 PM

How would you differentiate x+csc⁡xx+\csc{x}?

Below is the solution.



ddxx+csc⁡x\frac{d}{dx} x+\csc{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)).
(ddxx)+(ddxcsc⁡x)(\frac{d}{dx} x)+(\frac{d}{dx} \csc{x})

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

3
Use Trigonometric Differentiation: the derivative of csc⁡x\csc{x} is −csc⁡xcot⁡x-\csc{x}\cot{x}.
1−csc⁡xcot⁡x1-\csc{x}\cot{x}

Done
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