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f^(')(x) if
f(x)=(7)/(5)sin^(2)(2x)

$f^{\prime}(x)$ if $f(x)=\frac{7}{5} \sin ^{2}(2 x)$

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Relative maxima or minima

Full solution

Q.

f^{\prime}(x)

if

f(x)=\frac{7}{5} \sin ^{2}(2 x)

Identify function: First, identify the function to differentiate: $f(x) = \left(\frac{7}{5}\right)\sin^2(2x)$ .
Apply chain rule: Apply the chain rule for differentiation. Let $u = \sin(2x)$ , then $f(x) = \frac{7}{5}u^2$ . The derivative of $u^2$ is $2u \frac{du}{dx}$ .
Differentiate $u$ : Differentiate $u = \sin(2x)$ . Using the chain rule again, $\frac{du}{dx} = \cos(2x) \cdot$ derivative of $2x$ , which is $2$ . So, $\frac{du}{dx} = 2\cos(2x)$ .
Substitute back: Substitute back to find $f'(x)$ . $f'(x) = \frac{7}{5} \cdot 2 \cdot \sin(2x) \cdot 2\cos(2x)$ . Simplify to get $f'(x) = \frac{28}{5}\sin(2x)\cos(2x)$ .
Use double-angle identity: Use the double-angle identity for sine, $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ . Here, $\theta = 2x$ , so $\sin(4x) = 2\sin(2x)\cos(2x)$ . Therefore, $f'(x) = \frac{14}{5}\sin(4x)$ .

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