Question

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

Explanation:

Step1: Let $u = 3\pi t+7$ and $y=\sin^{2}u$.

Set up substitution.

Step2: First find $\frac{du}{dt}$.

$\frac{du}{dt}=\frac{d}{dt}(3\pi t + 7)=3\pi$

Step3: Then rewrite $y$ as $y = (\sin u)^{2}$ and find $\frac{dy}{du}$ using the chain - rule for $y = v^{2}$ and $v=\sin u$.

$\frac{dy}{dv}=2v$ and $\frac{dv}{du}=\cos u$, so $\frac{dy}{du}=\frac{dy}{dv}\cdot\frac{dv}{du}=2\sin u\cos u$

Step4: Use the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$.

Substitute $\frac{dy}{du}=2\sin u\cos u$ and $\frac{du}{dt}=3\pi$ into the chain - rule formula. Since $u = 3\pi t+7$, we have $\frac{dy}{dt}=2\sin(3\pi t + 7)\cos(3\pi t + 7)\cdot3\pi$

Step5: Simplify the result using the double - angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$.

$\frac{dy}{dt}=3\pi\sin(6\pi t + 14)$

Answer:

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