Problem of the Week

Updated at May 18, 2020 4:36 PM

For this week we've brought you this calculus problem.

How can we find the derivative of \(9q+\sin{q}\)?

Here are the steps:



\[\frac{d}{dq} 9q+\sin{q}\]

1
Use Sum Rule: \(\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))\).
\[(\frac{d}{dq} 9q)+(\frac{d}{dq} \sin{q})\]

2
Use Power Rule: \(\frac{d}{dx} {x}^{n}=n{x}^{n-1}\).
\[9+(\frac{d}{dq} \sin{q})\]

3
Use Trigonometric Differentiation: the derivative of \(\sin{x}\) is \(\cos{x}\).
\[9+\cos{q}\]

Done
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