Let g(x)=-8sin((3x)/(2)+1). Find g^('')(x). Choose 1 answer: (A) 8((3x)/(2)+1)^(2)*sin((3x)/(2)+1) (B) (32)/(9)sin((3x)/(2)+1) (C) 18 sin((3x)/(2)+1) (D) -12 cos((3x)/(2)+1)

Find First Derivative: Differentiate g(x) with respect to x to find the first derivative g'(x). g(x) = -8sin((3x)/(2)+1) Using the chain rule, the derivative of sin(u) with respect to x is cos(u) * du/dx, where u = (3x/2) + 1. g'(x) = -8 * cos((3x/2) + 1) * d/dx((3x/2) + 1) The derivative of (3x/2) + 1 with respect to x is 3/2. g'(x) = -8 * cos((3x/2) + 1) * (3/2) g'(x) = -12 * cos((3x/2) + 1)Calculate g'(x): Differentiate g'(x) to find the second derivative g''(x). g'(x) = -12 * cos((3x/2) + 1) Using the chain rule again, the derivative of cos(u) with respect to x is -sin(u) * du/dx, where u = (3x/2) + 1. g''(x) = -12 * (-sin((3x/2) + 1)) * d/dx((3x/2) + 1) The derivative of (3x/2) + 1 with respect to x is 3/2. g''(x) = -12 * (-sin((3x/2) + 1)) * (3/2) g''(x) = 18 * sin((3x/2) + 1)

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Let
g(x)=-8sin((3x)/(2)+1).
Find
g^('')(x).
Choose 1 answer:
(A)
8((3x)/(2)+1)^(2)*sin((3x)/(2)+1)
(B)
(32)/(9)sin((3x)/(2)+1)
(C)
18 sin((3x)/(2)+1)
(D)
-12 cos((3x)/(2)+1)

Let $g(x)=-8 \sin \left(\frac{3 x}{2}+1\right)$ . $\newline$ Find $g^{\prime \prime}(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $8\left(\frac{3 x}{2}+1\right)^{2} \cdot \sin \left(\frac{3 x}{2}+1\right)$ $\newline$ (B) $\frac{32}{9} \sin \left(\frac{3 x}{2}+1\right)$ $\newline$ (C) $18 \sin \left(\frac{3 x}{2}+1\right)$ $\newline$ (D) $-12 \cos \left(\frac{3 x}{2}+1\right)$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Let

g(x)=-8 \sin \left(\frac{3 x}{2}+1\right)

\newline

Find

g^{\prime \prime}(x)

\newline

Choose

1

answer:

\newline

(A)

8\left(\frac{3 x}{2}+1\right)^{2} \cdot \sin \left(\frac{3 x}{2}+1\right)

\newline

(B)

\frac{32}{9} \sin \left(\frac{3 x}{2}+1\right)

\newline

(C)

18 \sin \left(\frac{3 x}{2}+1\right)

\newline

(D)

-12 \cos \left(\frac{3 x}{2}+1\right)

Find First Derivative: Differentiate $g(x)$ with respect to $x$ to find the first derivative $g'(x)$ .
$g(x) = -8\sin\left(\frac{3x}{2}+1\right)$
Using the chain rule, the derivative of $\sin(u)$ with respect to $x$ is $\cos(u) \cdot \frac{du}{dx}$ , where $u = \frac{3x}{2} + 1$ .
$g'(x) = -8 \cdot \cos\left(\frac{3x}{2} + 1\right) \cdot \frac{d}{dx}\left(\frac{3x}{2} + 1\right)$
The derivative of $\frac{3x}{2} + 1$ with respect to $x$ is $x$ $1$ .
$x$ $2$
$x$ $3$
Calculate $g'(x)$ : Differentiate $g'(x)$ to find the second derivative $g''(x)$ .
$g'(x) = -12 \cdot \cos\left(\frac{3x}{2} + 1\right)$
Using the chain rule again, the derivative of $\cos(u)$ with respect to $x$ is $-\sin(u) \cdot \frac{du}{dx}$ , where $u = \frac{3x}{2} + 1$ .
$g''(x) = -12 \cdot (-\sin(\frac{3x}{2} + 1)) \cdot \frac{d}{dx}(\frac{3x}{2} + 1)$
The derivative of $\frac{3x}{2} + 1$ with respect to $x$ is $g'(x)$ $1$ .
$g'(x)$ $2$
$g'(x)$ $3$