find the derivative of the given function.\ny = cos (e^{-2\theta^{2}})\n\\frac{dy}{d\theta}= square

Explanation:

Step1: Apply chain - rule

Let $u = e^{-2\theta^{2}}$. Then $y=\cos(u)$. The chain - rule states that $\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}$. First, find $\frac{dy}{du}$. Since $y = \cos(u)$, $\frac{dy}{du}=-\sin(u)$.

Step2: Find $\frac{du}{d\theta}$

Let $v=-2\theta^{2}$. Then $u = e^{v}$. By the chain - rule, $\frac{du}{d\theta}=\frac{du}{dv}\cdot\frac{dv}{d\theta}$. Since $\frac{du}{dv}=e^{v}$ and $\frac{dv}{d\theta}=-4\theta$, then $\frac{du}{d\theta}=e^{-2\theta^{2}}\cdot(-4\theta)$.

Step3: Calculate $\frac{dy}{d\theta}$

Substitute $\frac{dy}{du}$ and $\frac{du}{d\theta}$ into the chain - rule formula $\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}$. We have $\frac{dy}{d\theta}=-\sin(e^{-2\theta^{2}})\cdot e^{-2\theta^{2}}\cdot(-4\theta)=4\theta e^{-2\theta^{2}}\sin(e^{-2\theta^{2}})$.

Answer:

$4\theta e^{-2\theta^{2}}\sin(e^{-2\theta^{2}})$