Find (d^(2))/(dx^(2))[2sin(-4x-3)].

Identify Function: Identify the function to differentiate. We are given the function f(x) = 2sin(-4x-3). We need to find the second derivative of this function with respect to x, which is denoted as (d^2)/(dx^2)[2sin(-4x-3)].Differentiate First Time: Differentiate the function with respect to x for the first time. The first derivative of f(x) with respect to x is found using the chain rule. The derivative of sin(u) with respect to u is cos(u), and the derivative of -4x-3 with respect to x is -4. Therefore, the first derivative f'(x) is: f'(x) = 2cos(-4x-3) * (-4) = -8cos(-4x-3).Differentiate Second Time: Differentiate the first derivative with respect to x for the second time. Now we need to find the second derivative, which is the derivative of f'(x). Again using the chain rule, the derivative of cos(u) with respect to u is -sin(u), and the derivative of -4x-3 with respect to x is -4. Therefore, the second derivative f''(x) is: f''(x) = -8(-sin(-4x-3)) * (-4) = -8 * 4 * sin(-4x-3) = -32sin(-4x-3).

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Q. Find

\frac{d^{2}}{d x^{2}}[2 \sin (-4 x-3)]

Identify Function: Identify the function to differentiate. $\newline$ We are given the function $f(x) = 2\sin(-4x-3)$ . We need to find the second derivative of this function with respect to $x$ , which is denoted as $\frac{d^2}{dx^2}[2\sin(-4x-3)]$ .
Differentiate First Time: Differentiate the function with respect to $x$ for the first time. $\newline$ The first derivative of $f(x)$ with respect to $x$ is found using the chain rule. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$ , and the derivative of $-4x-3$ with respect to $x$ is $-4$ . Therefore, the first derivative $f'(x)$ is: $\newline$ $f(x)$ $0$ .
Differentiate Second Time: Differentiate the first derivative with respect to $x$ for the second time. $\newline$ Now we need to find the second derivative, which is the derivative of $f'(x)$ . Again using the chain rule, the derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$ , and the derivative of $-4x-3$ with respect to $x$ is $-4$ . Therefore, the second derivative $f''(x)$ is: $\newline$ $f''(x) = -8(-\sin(-4x-3)) \times (-4) = -8 \times 4 \times \sin(-4x-3) = -32\sin(-4x-3)$ .