Question

find the derivative of the trigonometric function.
h(x) = sin(2x) cos(2x)
h(x) =

Explanation:

Step1: Apply product - rule

The product - rule states that if $h(x)=u(x)v(x)$, then $h'(x)=u'(x)v(x)+u(x)v'(x)$. Here, $u(x)=\sin(2x)$ and $v(x)=\cos(2x)$.

Step2: Find $u'(x)$

Using the chain - rule, if $y = \sin(u)$ and $u = 2x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{d}{du}\sin(u)=\cos(u)$ and $\frac{d}{dx}(2x) = 2$, we have $u'(x)=2\cos(2x)$.

Step3: Find $v'(x)$

Using the chain - rule, if $y=\cos(u)$ and $u = 2x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{d}{du}\cos(u)=-\sin(u)$ and $\frac{d}{dx}(2x)=2$, we have $v'(x)=-2\sin(2x)$.

Step4: Calculate $h'(x)$

Substitute $u(x),v(x),u'(x),v'(x)$ into the product - rule formula:

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Using the double - angle formula $\cos(2\alpha)=\cos^{2}\alpha-\sin^{2}\alpha$, we get $h'(x)=2\cos(4x)$.

Answer:

$2\cos(4x)$