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${:[f(x)=arcsin((x)/(3))],[f^(')(1)=◻]:} Use an exact expression.$

$\begin{array}{l} f(x)=\arcsin \left(\frac{x}{3}\right) \\ f^{\prime}(1)=\square \end{array}$ $\newline$ Use an exact expression.

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Find derivatives of inverse trigonometric functions

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Q.

\begin{array}{l} f(x)=\arcsin \left(\frac{x}{3}\right) \\ f^{\prime}(1)=\square \end{array}

\newline

Use an exact expression.

Apply Chain Rule: To find the derivative of $f(x) = \arcsin\left(\frac{x}{3}\right)$ , we need to use the chain rule. The derivative of $\arcsin(u)$ with respect to $u$ is $\frac{1}{\sqrt{1-u^2}}$ . Here, $u = \frac{x}{3}$ .
Derivative Calculation: Applying the chain rule, we get: $\newline$ $f'(x) = \frac{d}{dx}(\arcsin(\frac{x}{3})) = \frac{1}{\sqrt{1-(\frac{x}{3})^2}} \cdot \frac{d}{dx}(\frac{x}{3})$
Simplify Expression: The derivative of $\frac{x}{3}$ with respect to $x$ is $\frac{1}{3}$ . So, we have: $\newline$ $f'(x) = \left(\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}}\right) \cdot \frac{1}{3}$
Evaluate at $x=1$ : Simplifying the expression, we get: $f'(x) = \frac{1}{3\sqrt{1-\left(\frac{x}{3}\right)^2}}$
Calculate Inside Square Root: Now we need to evaluate the derivative at $x = 1$ . Plugging $x = 1$ into the derivative, we get: $\newline$ $f'(1) = \frac{1}{3\sqrt{1-(\frac{1}{3})^2}}$
Find $f'(1)$ : Calculating the value inside the square root, we have: $\newline$ $1 - (\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$
Find $f'(1)$ : Calculating the value inside the square root, we have: $1 - (\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$ Now, we can find the value of $f'(1)$ : $f'(1) = \frac{1}{3*\sqrt{\frac{8}{9}}} = \frac{1}{3*(\frac{2}{3})} = \frac{1}{3*\frac{2}{3}} = \frac{1}{2}$

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