find $\frac{dy}{dt}$. \n$y=(5 + cos 5t)^{-4}$\n$\frac{dy}{dt}=square$

Explanation:

Step1: Apply chain - rule

Let $u = 5+\cos5t$, then $y = u^{-4}$. The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$. Using the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$, we have $\frac{dy}{du}=-4u^{-5}$.

Step2: Find $\frac{du}{dt}$

Differentiate $u = 5+\cos5t$ with respect to $t$. The derivative of a constant is 0, and using the chain - rule for $\cos5t$. Let $v = 5t$, then $\frac{d}{dt}(\cos5t)=\frac{d}{dv}(\cos v)\cdot\frac{dv}{dt}$. Since $\frac{d}{dv}(\cos v)=-\sin v$ and $\frac{dv}{dt}=5$, we get $\frac{du}{dt}=-5\sin5t$.

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

Substitute $\frac{dy}{du}$ and $\frac{du}{dt}$ into the chain - rule formula $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. $\frac{dy}{dt}=-4u^{-5}\cdot(-5\sin5t)$. Replace $u = 5+\cos5t$ back in, we get $\frac{dy}{dt}=20(5 + \cos5t)^{-5}\sin5t$.

Answer:

$20\sin5t(5+\cos5t)^{-5}$