Let f(x)=2cos((x)/(2)). Find f^('')(x). Choose 1 answer: (A) -cos((x)/(2)) B -sin((x)/(2)) (C) -8cos((x)/(2)) (D) -(1)/(2)cos((x)/(2))

Differentiate function f(x): Differentiate the function f(x) = 2cos(x/2) with respect to x to find the first derivative f'(x). Using the chain rule, the derivative of cos(u) with respect to x is -sin(u) times the derivative of u with respect to x. Here, u = x/2, so the derivative of u with respect to x is 1/2. f'(x) = 2 * (-sin(x/2)) * (1/2)Simplify first derivative: Simplify the expression for the first derivative. f'(x) = -sin(x/2)Differentiate f'(x): Differentiate the first derivative f'(x) = -sin(x/2) with respect to x to find the second derivative f''(x). Using the chain rule again, the derivative of -sin(u) with respect to x is -cos(u) times the derivative of u with respect to x. Here, u = x/2, so the derivative of u with respect to x is 1/2. f''(x) = -cos(x/2) * (1/2)Simplify second derivative: Simplify the expression for the second derivative. f''(x) = -(1/2)cos(x/2)

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Let
f(x)=2cos((x)/(2)).
Find
f^('')(x).
Choose 1 answer:
(A)
-cos((x)/(2))
B
-sin((x)/(2))
(C)
-8cos((x)/(2))
(D)
-(1)/(2)cos((x)/(2))

Let $f(x)=2 \cos \left(\frac{x}{2}\right)$ . $\newline$ Find $f^{\prime \prime}(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-\cos \left(\frac{x}{2}\right)$ $\newline$ (B) $-\sin \left(\frac{x}{2}\right)$ $\newline$ (C) $-8 \cos \left(\frac{x}{2}\right)$ $\newline$ (D) $-\frac{1}{2} \cos \left(\frac{x}{2}\right)$

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Precalculus

Operations with rational exponents

Full solution

Q. Let

f(x)=2 \cos \left(\frac{x}{2}\right)

\newline

Find

f^{\prime \prime}(x)

\newline

Choose

1

answer:

\newline

(A)

-\cos \left(\frac{x}{2}\right)

\newline

(B)

-\sin \left(\frac{x}{2}\right)

\newline

(C)

-8 \cos \left(\frac{x}{2}\right)

\newline

(D)

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

Differentiate function $f(x)$ : Differentiate the function $f(x) = 2\cos(\frac{x}{2})$ with respect to $x$ to find the first derivative $f'(x)$ . Using the chain rule, the derivative of $\cos(u)$ with respect to $x$ is $-\sin(u)$ times the derivative of $u$ with respect to $x$ . Here, $u = \frac{x}{2}$ , so the derivative of $u$ with respect to $x$ is $f(x) = 2\cos(\frac{x}{2})$ $2$ . $f(x) = 2\cos(\frac{x}{2})$ $3$
Simplify first derivative: Simplify the expression for the first derivative. $f'(x) = -\sin(\frac{x}{2})$
Differentiate $f'(x)$ : Differentiate the first derivative $f'(x) = -\sin(\frac{x}{2})$ with respect to $x$ to find the second derivative $f''(x)$ . Using the chain rule again, the derivative of $-\sin(u)$ with respect to $x$ is $-\cos(u)$ times the derivative of $u$ with respect to $x$ . Here, $u = \frac{x}{2}$ , so the derivative of $u$ with respect to $x$ is $f'(x) = -\sin(\frac{x}{2})$ $2$ . $f'(x) = -\sin(\frac{x}{2})$ $3$
Simplify second derivative: Simplify the expression for the second derivative. $f''(x) = -(\frac{1}{2})\cos(\frac{x}{2})$