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 $\frac{dy}{du}$. Since $y = \cos(u)$, $\frac{dy}{du}=-\sin(u)$.

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

Since $u=\tan(3t - 2)$, and using the chain - rule again. Let $v = 3t-2$, then $u = \tan(v)$. The derivative of $\tan(v)$ with respect to $v$ is $\sec^{2}(v)$, and the derivative of $v = 3t - 2$ with respect to $t$ is 3. So $\frac{du}{dt}=\sec^{2}(v)\cdot3=3\sec^{2}(3t - 2)$.

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

Substitute $u=\tan(3t - 2)$ and $\frac{dy}{du}=-\sin(u)$ and $\frac{du}{dt}=3\sec^{2}(3t - 2)$ into the chain - rule formula $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. We get $\frac{dy}{dt}=-\sin(\tan(3t - 2))\cdot3\sec^{2}(3t - 2)=- 3\sec^{2}(3t - 2)\sin(\tan(3t - 2))$.

Answer:

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