Question

hw13 derivatives of trigonometric function
score: 1/7 answered: 1/7
question 2
if $f(x)=4sin x + 3cos x$, then $f(x)=$
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Explanation:

Step1: Recall derivative rules

The derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Also, use the sum - rule of derivatives $(u + v)'=u'+v'$.

Step2: Differentiate each term

For $u = 4\sin x$, by the constant - multiple rule $(cf(x))'=cf'(x)$ where $c = 4$ and $f(x)=\sin x$, we have $u'=4\cos x$. For $v = 3\cos x$, by the constant - multiple rule with $c = 3$ and $f(x)=\cos x$, we have $v'=- 3\sin x$.

Step3: Apply sum - rule

$f'(x)=u'+v'=4\cos x-3\sin x$.

Answer:

$4\cos x - 3\sin x$