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 \(3z+\ln{z}\)?

Here are the steps:



\[\frac{d}{dz} 3z+\ln{z}\]

1
Use Sum Rule: \(\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))\).
\[(\frac{d}{dz} 3z)+(\frac{d}{dz} \ln{z})\]

2
Use Power Rule: \(\frac{d}{dx} {x}^{n}=n{x}^{n-1}\).
\[3+(\frac{d}{dz} \ln{z})\]

3
The derivative of \(\ln{x}\) is \(\frac{1}{x}\).
\[3+\frac{1}{z}\]

Done
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