# 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 $$\frac{1}{x}$$.$3+\frac{1}{z}$Done3+1/z