The first derivative of the function (sin 2x | vedclass

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(B) Let $f(x) = \sin 2x \cos 2x \cos 3x + \log_2 2^{x+3}$.Using the identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$ and the property $\log_a a^k = k$

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$-1$

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(B) Let $f(x) = \sin 2x \cos 2x \cos 3x + \log_2 2^{x+3}$.Using the identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$ and the property $\log_a a^k = k$,we get:$f(x) = \frac{1}{2} \sin 4x \cos 3x + (x + 3)$.Using the product-to-sum formula $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$,we have:$f(x) = \frac{1}{2} \cdot \frac{1}{2} [\sin(4x+3x) + \sin(4x-3x)] + x + 3$$f(x) = \frac{1}{4} [\sin 7x + \sin x] + x + 3$.Differentiating with respect to $x$:$f'(x) = \frac{1}{4} [7 \cos 7x + \cos x] + 1$.Evaluating at $x = \pi$:$f'(\pi) = \frac{1}{4} [7 \cos(7\pi) + \cos(\pi)] + 1$Since $\cos(7\pi) = -1$ and $\cos(\pi) = -1$:$f'(\pi) = \frac{1}{4} [7(-1) + (-1)] + 1$$f'(\pi) = \frac{1}{4} [-8] + 1 = -2 + 1 = -1$.

The first derivative of the function $(\sin 2x \cos 2x \cos 3x + \log_2 2^{x+3})$ with respect to $x$ at $x = \pi$ is

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