# Problem of the Week

## Updated at Mar 13, 2023 12:34 PM

To get more practice in calculus, we brought you this problem of the week:

How would you differentiate $$\sqrt{q}+\ln{q}$$?

Check out the solution below!

$\frac{d}{dq} \sqrt{q}+\ln{q}$

 1 Use Sum Rule: $$\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))$$.$(\frac{d}{dq} \sqrt{q})+(\frac{d}{dq} \ln{q})$2 Since $$\sqrt{x}={x}^{\frac{1}{2}}$$, using the Power Rule, $$\frac{d}{dx} {x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}}$$$\frac{1}{2\sqrt{q}}+(\frac{d}{dq} \ln{q})$3 The derivative of $$\ln{x}$$ is $$\frac{1}{x}$$.$\frac{1}{2\sqrt{q}}+\frac{1}{q}$Done1/(2*sqrt(q))+1/q