Problem of the Week

Updated at May 2, 2016 2:17 PM

How can we solve for the derivative of cot⁡x−ln⁡x\cot{x}-\ln{x}?

Below is the solution.



ddxcot⁡x−ln⁡x\frac{d}{dx} \cot{x}-\ln{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)).
(ddxcot⁡x)−(ddxln⁡x)(\frac{d}{dx} \cot{x})-(\frac{d}{dx} \ln{x})

2
Use Trigonometric Differentiation: the derivative of cot⁡x\cot{x} is −csc⁡2x-\csc^{2}x.
−csc⁡2x−(ddxln⁡x)-\csc^{2}x-(\frac{d}{dx} \ln{x})

3
The derivative of ln⁡x\ln{x} is 1x\frac{1}{x}.
−csc⁡2x−1x-\csc^{2}x-\frac{1}{x}

Done
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