g(x)=arccos(2x) g^(')(x)=? Choose 1 answer: (A) (-2)/(sqrt(1-4x^(2))) (B) (2)/(sqrt(1-4x^(2))) (C) (-2)/(sqrt(4x^(2)-1)) (D) (2)/(sqrt(4x^(2)-1))

Chain Rule Differentiation: step_1: Use the chain rule to differentiate g(x) = arccos(2x). The derivative of arccos(u) with respect to u is -1/sqrt(1-u^2). Let u = 2x, then du/dx = 2. g'(x) = d/dx(arccos(u)) * du/dx = (-1/sqrt(1-u^2)) * (2).Substitute u into Derivative: step_2: Substitute u = 2x into the derivative. g'(x) = (-1/sqrt(1-(2x)^2)) * (2) = (-2)/(sqrt(1-4x^2)).

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g(x)=arccos(2x)

g^(')(x)=?
Choose 1 answer:
(A)
(-2)/(sqrt(1-4x^(2)))
(B)
(2)/(sqrt(1-4x^(2)))
(C)
(-2)/(sqrt(4x^(2)-1))
(D)
(2)/(sqrt(4x^(2)-1))

$g(x)=\arccos (2 x)$ $\newline$ $g^{\prime}(x)=?$ $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{-2}{\sqrt{1-4 x^{2}}}$ $\newline$ (B) $\frac{2}{\sqrt{1-4 x^{2}}}$ $\newline$ (C) $\frac{-2}{\sqrt{4 x^{2}-1}}$ $\newline$ (D) $\frac{2}{\sqrt{4 x^{2}-1}}$

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Math Problems

Calculus

Find derivatives of inverse trigonometric functions

Full solution

g(x)=\arccos (2 x)

\newline

g^{\prime}(x)=?

\newline

Choose

1

answer:

\newline

(A)

\frac{-2}{\sqrt{1-4 x^{2}}}

\newline

(B)

\frac{2}{\sqrt{1-4 x^{2}}}

\newline

(C)

\frac{-2}{\sqrt{4 x^{2}-1}}

\newline

(D)

\frac{2}{\sqrt{4 x^{2}-1}}

Chain Rule Differentiation: step_1: Use the chain rule to differentiate $g(x) = \arccos(2x)$ . The derivative of $\arccos(u)$ with respect to $u$ is $-\frac{1}{\sqrt{1-u^2}}$ . Let $u = 2x$ , then $\frac{du}{dx} = 2$ . $g'(x) = \frac{d}{dx}(\arccos(u)) \cdot \frac{du}{dx} = \left(-\frac{1}{\sqrt{1-u^2}}\right) \cdot (2)$ .
Substitute u into Derivative: step_2: Substitute $u = 2x$ into the derivative. $g'(x) = (-1/\sqrt{1-(2x)^2}) \times (2) = (-2)/(\sqrt{1-4x^2})$ .