Problem of the Week

Updated at May 18, 2020 4:36 PM

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May 18, 2020

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

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

Here are the steps:

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

Use Sum Rule: \(\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x))\).

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

Use Power Rule: \(\frac{d}{dx} {x}^{n}=n{x}^{n-1}\).

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

Use Trigonometric Differentiation: the derivative of \(\sin{x}\) is \(\cos{x}\).

\[9+\cos{q}\]

Done