Question

find $\frac{dy}{dt}$.
$y = cos(\tan(3t - 2))$
$\frac{dy}{dt}=square$

Explanation:

Step1: Apply chain - rule

Let $u = \tan(3t - 2)$, then $y=\cos(u)$. The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$.
$\frac{dy}{du}=-\sin(u)$

Step2: Find $\frac{du}{dt}$

Let $v = 3t-2$, then $u = \tan(v)$. By the chain - rule, $\frac{du}{dt}=\frac{du}{dv}\cdot\frac{dv}{dt}$. We know that $\frac{du}{dv}=\sec^{2}(v)$ and $\frac{dv}{dt}=3$. So $\frac{du}{dt}=3\sec^{2}(3t - 2)$

Step3: Calculate $\frac{dy}{dt}$

Substitute $u = \tan(3t - 2)$ into $\frac{dy}{du}$ and multiply by $\frac{du}{dt}$:
$\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}=-\sin(\tan(3t - 2))\cdot3\sec^{2}(3t - 2)=- 3\sec^{2}(3t - 2)\sin(\tan(3t - 2))$

Answer:

$-3\sec^{2}(3t - 2)\sin(\tan(3t - 2))$