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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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Find derivatives of inverse trigonometric functions

Full solution

Q.

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. $g'(x) = (-1/\sqrt{1-(2x)^2}) \times (2) = (-2)/(\sqrt{1-4x^2})$ .

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