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⁡q9q+\sin{q}?

Here are the steps:



ddq9q+sin⁡q\frac{d}{dq} 9q+\sin{q}

1
Use Sum Rule: ddxf(x)+g(x)=(ddxf(x))+(ddxg(x))\frac{d}{dx} f(x)+g(x)=(\frac{d}{dx} f(x))+(\frac{d}{dx} g(x)).
(ddq9q)+(ddqsin⁡q)(\frac{d}{dq} 9q)+(\frac{d}{dq} \sin{q})

2
Use Power Rule: ddxxn=nxn−1\frac{d}{dx} {x}^{n}=n{x}^{n-1}.
9+(ddqsin⁡q)9+(\frac{d}{dq} \sin{q})

3
Use Trigonometric Differentiation: the derivative of sin⁡x\sin{x} is cos⁡x\cos{x}.
9+cos⁡q9+\cos{q}

Done
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