Problem of the Week

Updated at Dec 4, 2023 8:07 AM

How would you differentiate csc⁡n+ln⁡n\csc{n}+\ln{n}?

Below is the solution.



ddncsc⁡n+ln⁡n\frac{d}{dn} \csc{n}+\ln{n}

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)).
(ddncsc⁡n)+(ddnln⁡n)(\frac{d}{dn} \csc{n})+(\frac{d}{dn} \ln{n})

2
Use Trigonometric Differentiation: the derivative of csc⁡x\csc{x} is −csc⁡xcot⁡x-\csc{x}\cot{x}.
−csc⁡ncot⁡n+(ddnln⁡n)-\csc{n}\cot{n}+(\frac{d}{dn} \ln{n})

3
The derivative of ln⁡x\ln{x} is 1x\frac{1}{x}.
−csc⁡ncot⁡n+1n-\csc{n}\cot{n}+\frac{1}{n}

Done
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