Question

find $\frac{dy}{dt}$. $y=(tsec t)^{6}$ $\frac{dy}{dt}=square$

Explanation:

Step1: Apply chain - rule

Let $u = t\sec t$, then $y = u^{6}$. By the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=6u^{5}=6(t\sec t)^{5}$.

Step2: Apply product - rule to find $\frac{du}{dt}$

Since $u = t\sec t$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = t$ and $v=\sec t$. We know that $\frac{d}{dt}(t) = 1$ and $\frac{d}{dt}(\sec t)=\sec t\tan t$. So $\frac{du}{dt}=\sec t+t\sec t\tan t$.

Step3: Calculate $\frac{dy}{dt}$

$\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}=6(t\sec t)^{5}(\sec t + t\sec t\tan t)$.

Answer:

$6(t\sec t)^{5}(\sec t + t\sec t\tan t)$