Question

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find $\frac{dy}{dt}$.
y = (1 + cos 4t)^{-5}
$\frac{dy}{dt}=square$

Explanation:

Step1: Apply chain - rule

Let $u = 1+\cos4t$, then $y = u^{-5}$. 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 $\frac{dy}{du}=-5u^{-6}$.

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

Since $u = 1+\cos4t$, then $\frac{du}{dt}=\frac{d}{dt}(1)+\frac{d}{dt}(\cos4t)$. The derivative of a constant $\frac{d}{dt}(1) = 0$. For $\frac{d}{dt}(\cos4t)$, let $v = 4t$, then $\frac{d}{dt}(\cos4t)=\frac{d(\cos v)}{dv}\cdot\frac{dv}{dt}$. We know that $\frac{d(\cos v)}{dv}=-\sin v$ and $\frac{dv}{dt}=4$. So $\frac{d}{dt}(\cos4t)=-4\sin4t$. Then $\frac{du}{dt}=-4\sin4t$.

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

By the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Substitute $\frac{dy}{du}=-5u^{-6}$ and $\frac{du}{dt}=-4\sin4t$ into it. Replace $u = 1+\cos4t$ back in, we get $\frac{dy}{dt}=(-5(1 + \cos4t)^{-6})\cdot(-4\sin4t)$.
Simplify the expression: $\frac{dy}{dt}=20\sin4t(1 + \cos4t)^{-6}=\frac{20\sin4t}{(1+\cos4t)^{6}}$.

Answer:

$\frac{20\sin4t}{(1 + \cos4t)^{6}}$