Question

find $\frac{dy}{dt}$ for $y = sin^{2}(5pi t - 2)$. thats incorrect. correct answer: $10pisin(5pi t - 2)cos(5pi t - 2)$ your answer: $5picos(10pi t - 4)$ similar question ok

Explanation:

Step1: Let $u = \sin(5\pi t - 2)$.

Then $y = u^{2}$.

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

$\frac{dy}{du}=2u$

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

Using the chain - rule, if $u=\sin(5\pi t - 2)$, let $v = 5\pi t-2$. Then $\frac{du}{dv}=\cos v$ and $\frac{dv}{dt}=5\pi$. So $\frac{du}{dt}=\frac{du}{dv}\cdot\frac{dv}{dt}=\cos(5\pi t - 2)\cdot5\pi = 5\pi\cos(5\pi t - 2)$

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

Substitute $\frac{dy}{du}=2u$ and $\frac{du}{dt}=5\pi\cos(5\pi t - 2)$ into the chain - rule formula. Since $u = \sin(5\pi t - 2)$, we have $\frac{dy}{dt}=2\sin(5\pi t - 2)\cdot5\pi\cos(5\pi t - 2)=10\pi\sin(5\pi t - 2)\cos(5\pi t - 2)$

Answer:

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