Question

find $\frac{dy}{dt}$ for $y = cos^{2}(5pi t + 1)$.

Explanation:

Step1: Let $u = 5\pi t+1$ and $y=\cos^{2}u$.

Set up substitution.

Step2: First find $\frac{dy}{du}$ using chain - rule for $y = (\cos u)^2$.

Let $v=\cos u$, then $y = v^{2}$. $\frac{dy}{dv}=2v$ and $\frac{dv}{du}=-\sin u$. So $\frac{dy}{du}=\frac{dy}{dv}\cdot\frac{dv}{du}=2\cos u(-\sin u)=-\sin(2u)$ (using double - angle formula $\sin(2\alpha) = 2\sin\alpha\cos\alpha$).

Step3: Then find $\frac{du}{dt}$.

Since $u = 5\pi t + 1$, $\frac{du}{dt}=5\pi$.

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

Substitute $\frac{dy}{du}=-\sin(2u)$ and $\frac{du}{dt}=5\pi$ into the chain - rule formula. Replace $u = 5\pi t+1$, we get $\frac{dy}{dt}=- 5\pi\sin(2(5\pi t + 1))=-5\pi\sin(10\pi t+2)$.

Answer:

$-5\pi\sin(10\pi t + 2)$