find the derivative of the trigonometric function.\nf(\theta)=\frac{1}{4}sin^{2}(2\theta)\nf(\theta)=

Explanation:

Step1: Apply chain - rule

Let $u = \sin(2\theta)$. Then $f(\theta)=\frac{1}{4}u^{2}$. First, find $\frac{df}{du}$ and $\frac{du}{d\theta}$. The derivative of $y = \frac{1}{4}u^{2}$ with respect to $u$ is $\frac{df}{du}=\frac{1}{4}\times2u=\frac{1}{2}u$.

Step2: Differentiate $u = \sin(2\theta)$

Using the chain - rule again, let $v = 2\theta$. Then $u=\sin(v)$. The derivative of $u$ with respect to $v$ is $\frac{du}{dv}=\cos(v)$, and the derivative of $v$ with respect to $\theta$ is $\frac{dv}{d\theta}=2$. So, $\frac{du}{d\theta}=\frac{du}{dv}\cdot\frac{dv}{d\theta}=\cos(2\theta)\times2 = 2\cos(2\theta)$.

Step3: Use the chain - rule for $f(\theta)$

By the chain - rule $\frac{df}{d\theta}=\frac{df}{du}\cdot\frac{du}{d\theta}$. Substitute $u = \sin(2\theta)$ and $\frac{du}{d\theta}=2\cos(2\theta)$ into the formula. We get $\frac{df}{d\theta}=\frac{1}{2}\sin(2\theta)\times2\cos(2\theta)$.

Step4: Simplify the result

$\frac{df}{d\theta}=\sin(2\theta)\cos(2\theta)$.

Answer:

$\sin(2\theta)\cos(2\theta)$