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 \(\frac{1}{x}\).
\[\frac{1}{2\sqrt{q}}+\frac{1}{q}\]

Done
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