Question

find $\frac{dy}{dt}$. $y = e^{cos^{2}(pi t - 2)}$ $\frac{dy}{dt}=square$

Explanation:

Step1: Apply chain - rule

Let $u = \cos^{2}(\pi t - 2)$. Then $y = e^{u}$. By the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$. Since $y = e^{u}$, $\frac{dy}{du}=e^{u}$.

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

Let $v=\cos(\pi t - 2)$. Then $u = v^{2}$. By the chain - rule $\frac{du}{dt}=\frac{du}{dv}\cdot\frac{dv}{dt}$. Since $u = v^{2}$, $\frac{du}{dv}=2v$. And since $v=\cos(\pi t - 2)$, by the chain - rule $\frac{dv}{dt}=-\sin(\pi t - 2)\cdot\pi$.

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

Substitute $v=\cos(\pi t - 2)$ into $\frac{du}{dt}=\frac{du}{dv}\cdot\frac{dv}{dt}$. We get $\frac{du}{dt}=2\cos(\pi t - 2)\cdot(-\pi\sin(\pi t - 2))=-\pi\sin(2(\pi t - 2))$ (using the double - angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$).

Step4: Calculate $\frac{dy}{dt}$

Substitute $u = \cos^{2}(\pi t - 2)$ and $\frac{du}{dt}=-\pi\sin(2(\pi t - 2))$ into $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. We have $\frac{dy}{dt}=e^{\cos^{2}(\pi t - 2)}\cdot(-\pi\sin(2(\pi t - 2)))$.

Answer:

$-\pi\sin(2(\pi t - 2))e^{\cos^{2}(\pi t - 2)}$