Question

find $\frac{dy}{dt}$ for $y = sin^{2}(3pi t + 7)$.

Explanation:

Step1: Let $u = 3\pi t+7$

$y=\sin^{2}u$

Step2: Apply chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$

First find $\frac{dy}{du}$. Let $y = v^{2}$ and $v=\sin u$. Then $\frac{dy}{dv} = 2v$ and $\frac{dv}{du}=\cos u$. So $\frac{dy}{du}=\frac{dy}{dv}\cdot\frac{dv}{du}=2\sin u\cos u=\sin(2u)$ by the double - angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$.

Step3: Find $\frac{du}{dt}$

Since $u = 3\pi t + 7$, then $\frac{du}{dt}=3\pi$.

Step4: Calculate $\frac{dy}{dt}$

$\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}=\sin(2u)\cdot3\pi$. Substitute $u = 3\pi t+7$ back in, we get $\frac{dy}{dt}=3\pi\sin(2(3\pi t + 7))=3\pi\sin(6\pi t+14)$.

Answer:

$3\pi\sin(6\pi t + 14)$