Question

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

Explanation:

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

$y = u^{2}$

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

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

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

$\frac{du}{dt}=\sec(6\pi t)\tan(6\pi t)\cdot6\pi$

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

$\frac{dy}{dt}=2u\cdot\sec(6\pi t)\tan(6\pi t)\cdot6\pi$

Step5: Substitute $u = \sec(6\pi t)$ back in

$\frac{dy}{dt}=2\sec(6\pi t)\cdot\sec(6\pi t)\tan(6\pi t)\cdot6\pi=12\pi\sec^{2}(6\pi t)\tan(6\pi t)$

Answer:

$12\pi\sec^{2}(6\pi t)\tan(6\pi t)$