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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\log_a a^k = k$

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(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

Similar Questions

Differentiate the function with respect to $x$: $\cos (\sqrt{x})$

If $y = \tan^{-1}(\frac{1}{x^2+x+1}) + \tan^{-1}(\frac{1}{x^2+3x+3}) + \tan^{-1}(\frac{1}{x^2+5x+7}) + \dots$ upto $n$ terms, then $y'(0) = $

The derivative of $f(x) = x|x|$ is

If $y = e^{\sqrt{x}}$,then $\frac{dy}{dx}$ equals:

Let $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$,where $x \in R$. Then $f'(10)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module