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y=arcsin((x)/(4))

(dy)/(dx)=?
Choose 1 answer:
(A)
(1)/(sqrt(1-(x^(2))/(16)))
(B)
(1)/(4sqrt(1-(x^(2))/(16)))
(c)
(1)/(sqrt(1-(x^(2))/(4)))
(D)
(1)/(4sqrt(1-(x^(2))/(4)))

$y=\arcsin \left(\frac{x}{4}\right)$ $\newline$ $\frac{d y}{d x}=?$ $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{\sqrt{1-\frac{x^{2}}{16}}}$ $\newline$ (B) $\frac{1}{4 \sqrt{1-\frac{x^{2}}{16}}}$ $\newline$ (c) $\frac{1}{\sqrt{1-\frac{x^{2}}{4}}}$ $\newline$ (D) $\frac{1}{4 \sqrt{1-\frac{x^{2}}{4}}}$

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Find derivatives of inverse trigonometric functions

Full solution

Q.

y=\arcsin \left(\frac{x}{4}\right)

\newline

\frac{d y}{d x}=?

\newline

Choose

1

answer:

\newline

(A)

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

\newline

(B)

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

\newline

(c)

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

\newline

(D)

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

Apply Chain Rule: To find the derivative of $y = \arcsin\left(\frac{x}{4}\right)$ , we use the chain rule.
Derivative of arcsin(u): The derivative of $\text{arcsin}(u)$ with respect to $u$ is $\frac{1}{\sqrt{1-u^2}}$ . Here, $u = \frac{x}{4}$ .
Derivative of $u$ : Now we need to find the derivative of $u = \frac{x}{4}$ with respect to $x$ , which is $\frac{1}{4}$ .
Apply Chain Rule Again: Applying the chain rule, the derivative of $y$ with respect to $x$ is $\frac{1}{\sqrt{1-(\frac{x}{4})^2}}$ * $\frac{du}{dx}$ .
Substitute $\frac{du}{dx}$ : Substitute $\frac{du}{dx} = \frac{1}{4}$ into the equation, we get $\frac{1}{\sqrt{1-(\frac{x}{4})^2}} \times \frac{1}{4}$ .
Simplify Final Answer: Simplify the expression to get the final answer: $(\frac{1}{4\sqrt{1-(\frac{x^2}{16})}})$ .

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