Question

find the derivative of $y = 3cos(3x + 1)$

Explanation:

Step1: Identify outer - inner functions

Let $u = 3x+1$, then $y = 3\cos(u)$.

Step2: Differentiate outer function

The derivative of $y$ with respect to $u$, $\frac{dy}{du}=- 3\sin(u)$.

Step3: Differentiate inner function

The derivative of $u$ with respect to $x$, $\frac{du}{dx}=3$.

Step4: Apply chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the formula: $\frac{dy}{dx}=-3\sin(u)\cdot3$.

Step5: Substitute back $u$

Since $u = 3x + 1$, we have $\frac{dy}{dx}=-9\sin(3x + 1)$.

Answer:

$-9\sin(3x + 1)$