Question

find the derivative of the given function.
y = cos(e^{-5\theta^{2}})
$\frac{dy}{d\theta}=square$

Explanation:

Step1: Apply chain - rule

Let $u = e^{-5\theta^{2}}$. 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 $v=-5\theta^{2}$. Then $u = e^{v}$. By the chain - rule again, $\frac{du}{d\theta}=\frac{du}{dv}\cdot\frac{dv}{d\theta}$. Since $\frac{du}{dv}=e^{v}$ and $\frac{dv}{d\theta}=-10\theta$, then $\frac{du}{d\theta}=e^{-5\theta^{2}}\cdot(- 10\theta)$.

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

Substitute $\frac{dy}{du}$ and $\frac{du}{d\theta}$ into the chain - rule formula. $\frac{dy}{d\theta}=-\sin(u)\cdot\frac{du}{d\theta}$. Replace $u = e^{-5\theta^{2}}$ and $\frac{du}{d\theta}=e^{-5\theta^{2}}\cdot(-10\theta)$. So $\frac{dy}{d\theta}=(-\sin(e^{-5\theta^{2}}))\cdot(-10\theta e^{-5\theta^{2}})=10\theta e^{-5\theta^{2}}\sin(e^{-5\theta^{2}})$.

Answer:

$10\theta e^{-5\theta^{2}}\sin(e^{-5\theta^{2}})$