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${:[h(x)=arcsin(4x)],[h^(')(x)=?]:} Choose 1 answer: (A) (1)/(sqrt(1-16x^(2))) (B) (4)/(sqrt(1-16x^(2))) (C) (1)/(sqrt(1+16x^(2))) (D) (4)/(sqrt(1+16x^(2)))$

$\begin{array}{l} h(x)=\arcsin (4 x) \\ h^{\prime}(x)=? \end{array}$ $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{\sqrt{1-16 x^{2}}}$ $\newline$ (B) $\frac{4}{\sqrt{1-16 x^{2}}}$ $\newline$ (C) $\frac{1}{\sqrt{1+16 x^{2}}}$ $\newline$ (D) $\frac{4}{\sqrt{1+16 x^{2}}}$

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Calculus

Find derivatives of inverse trigonometric functions

Full solution

\begin{array}{l} h(x)=\arcsin (4 x) \\ h^{\prime}(x)=? \end{array}

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{\sqrt{1-16 x^{2}}}

\newline

(B)

\frac{4}{\sqrt{1-16 x^{2}}}

\newline

(C)

\frac{1}{\sqrt{1+16 x^{2}}}

\newline

(D)

\frac{4}{\sqrt{1+16 x^{2}}}

Apply Chain Rule: To find the derivative of $h(x) = \arcsin(4x)$ , we use the chain rule.
Derivative of arcsin(u): The derivative of $\arcsin(u)$ with respect to $u$ is $\frac{1}{\sqrt{1-u^2}}$ . Here, $u = 4x$ .
Calculate $\frac{du}{dx}$ : Using the chain rule, $h'(x) = \frac{d}{dx}(\arcsin(4x)) = \frac{d}{dx}(\arcsin(u)) \cdot \frac{du}{dx}$ .
Plug in Derivatives: We calculate $\frac{du}{dx}$ where $u = 4x$ . So, $\frac{du}{dx} = 4$ .
Simplify Expression: Now, plug in the derivative of $u$ and the derivative of $\arcsin(u)$ into the chain rule formula. $\newline$ $h'(x) = \frac{1}{\sqrt{1-(4x)^2}} \times 4$ .
Simplify Expression: Now, plug in the derivative of $u$ and the derivative of $\arcsin(u)$ into the chain rule formula. $\newline$ $h'(x) = \frac{1}{\sqrt{1-(4x)^2}} \times 4$ . Simplify the expression to get the final derivative. $\newline$ $h'(x) = \frac{4}{\sqrt{1-16x^2}}$ .