Problem of the Week

Updated at Nov 16, 2020 9:11 AM

For this week we've brought you this calculus problem.

How can we solve for the derivative of 3z+ln⁡z3z+\ln{z}?

Here are the steps:



ddz3z+ln⁡z\frac{d}{dz} 3z+\ln{z}

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)).
(ddz3z)+(ddzln⁡z)(\frac{d}{dz} 3z)+(\frac{d}{dz} \ln{z})

2
Use Power Rule: ddxxn=nxn−1\frac{d}{dx} {x}^{n}=n{x}^{n-1}.
3+(ddzln⁡z)3+(\frac{d}{dz} \ln{z})

3
The derivative of ln⁡x\ln{x} is 1x\frac{1}{x}.
3+1z3+\frac{1}{z}

Done
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