Question

find $\frac{dy}{dt}$ for $y = sin^{2}(5pi t - 2)$. $\frac{dy}{dt}=5picos(10pi t - 4)$

Explanation:

Step1: Apply chain - rule

Let $u = 5\pi t-2$, then $y=\sin^{2}u$. First find $\frac{dy}{du}$ and $\frac{du}{dt}$.

Step2: Differentiate $y$ with respect to $u$

Using the power - rule and the derivative of sine function, if $y = \sin^{2}u$, then $\frac{dy}{du}=2\sin u\cos u=\sin(2u)$ (by double - angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$).

Step3: Differentiate $u$ with respect to $t$

Since $u = 5\pi t - 2$, $\frac{du}{dt}=5\pi$.

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

Substitute $u = 5\pi t-2$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dt}$. We get $\frac{dy}{dt}=\sin(2(5\pi t - 2))\cdot5\pi=5\pi\cos(10\pi t - 4)$ (using the identity $\sin(2\alpha)=\cos(\frac{\pi}{2}-2\alpha)$ or just the chain - rule result).

Answer:

$5\pi\cos(10\pi t - 4)$