Question

question 29 find the derivative of $y = \sin(3x^2 - 4x + 1)$. \\(\frac{dy}{dx} = (6x - 4)\cos(3x^2 - 4x + 1)\\) \\(\frac{dy}{dx} = -\cos(6x - 4)\\) \\(\frac{dy}{dx} = - (6x - 4)\cos(3x^2 - 4x + 1)\\) \\(\frac{dy}{dx} = \cos(6x - 4)\\)

Explanation:

Step1: Identify inner/outer functions

Let $u=3x^2-4x+1$, $y=\sin(u)$.

Step2: Derive outer function

$\frac{dy}{du}=\cos(u)$

Step3: Derive inner function

$\frac{du}{dx}=6x-4$

Step4: Apply chain rule

$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}$
Substitute: $\frac{dy}{dx}=\cos(3x^2-4x+1) \cdot (6x-4)$

Answer:

$\frac{dy}{dx}=(6x-4)\cos(3x^2-4x+1)$