Question

question given the function $f(x)=-4cos(3x)$, find $f(x)$. answer attempt 1 out of 2 $f(x)=$

Explanation:

Step1: Recall chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Here $y=-4\cos(3x)$, let $u = 3x$, so $y=-4\cos(u)$.

Step2: Differentiate outer function

The derivative of $y=-4\cos(u)$ with respect to $u$ is $y'_u = 4\sin(u)$ (since the derivative of $\cos(u)$ is $-\sin(u)$ and the constant $- 4$ is multiplied).

Step3: Differentiate inner function

The derivative of $u = 3x$ with respect to $x$ is $u'_x=3$.

Step4: Apply chain - rule

By the chain - rule $f'(x)=y'_u\cdot u'_x$. Substitute $y'_u = 4\sin(u)$ and $u'_x = 3$ back in, and replace $u$ with $3x$. We get $f'(x)=4\sin(3x)\cdot3$.

Answer:

$12\sin(3x)$