Question

How do you find the derivative of $y=\cos \left(3x\right)$ ?

Answer 1

Solve for differentiation:

Given: $\frac{dy}{dx}=\frac{d}{dx}\left(\cos \right(3x\left)\right)$
we know that $\frac{d}{dx}\left(\cos \right(x\left)\right)=-\sin \left(x\right)$

applying chain rule $\frac{df\left(g\right(x\left)\right)}{dx}=\frac{d}{d\left(g\right(x\left)\right)}f\left(g\right(x\left)\right)\times \frac{dg\left(x\right)}{dx}$

Here, $g\left(x\right)=3x$ and $f\left(g\right(x\left)\right)=\cos \left(3x\right)$

$\frac{d}{dx}\left(\cos \right(3x\left)\right)=\frac{d}{d\left(3x\right)}\left(\cos \right(3x\left)\right)\times \frac{d}{dx}\left(3x\right)$

$\frac{dy}{dx}=(-\sin (3x\left)\right)\frac{d}{dx}\left(3x\right)$

$$\begin{gathered} \frac{dy}{dx}=(-\sin (3x\left)\right)3 \\ \frac{dy}{dx}=-3\sin \left(3x\right) \end{gathered}$$

Hence, the derivative of $y=\cos \left(3x\right)$ is $\frac{dy}{dx}=-3\sin \left(3x\right)$.