# Problem of the Week

## Updated at Jul 12, 2021 10:27 AM

How can we find the derivative of $$8u+\ln{u}$$?

Below is the solution.

$\frac{d}{du} 8u+\ln{u}$

 1 Use Sum Rule: $$\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))$$.$(\frac{d}{du} 8u)+(\frac{d}{du} \ln{u})$2 Use Power Rule: $$\frac{d}{dx} {x}^{n}=n{x}^{n-1}$$.$8+(\frac{d}{du} \ln{u})$3 The derivative of $$\ln{x}$$ is $$\frac{1}{x}$$.$8+\frac{1}{u}$Done8+1/u