Question

find $\frac{dy}{dt}$. $y = csc^{2}6pi t$ $\frac{dy}{dt}=square$

Explanation:

Step1: Let $u = \csc(6\pi t)$

$y = u^{2}$

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

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

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

$u=\csc(6\pi t)$, using the chain - rule, if $u = \csc(v)$ and $v = 6\pi t$, then $\frac{du}{dv}=-\csc(v)\cot(v)$ and $\frac{dv}{dt}=6\pi$. So $\frac{du}{dt}=-\csc(6\pi t)\cot(6\pi t)\cdot6\pi=- 6\pi\csc(6\pi t)\cot(6\pi t)$

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

Substitute $\frac{dy}{du}=2u$ and $\frac{du}{dt}=-6\pi\csc(6\pi t)\cot(6\pi t)$ into the chain - rule formula. Since $u = \csc(6\pi t)$, we have $\frac{dy}{dt}=2\csc(6\pi t)\cdot(-6\pi\csc(6\pi t)\cot(6\pi t))$

Answer:

$- 12\pi\csc^{2}(6\pi t)\cot(6\pi t)$