$y=\arccos \left(-\frac{x}{5}\right)$ $\newline$ Evaluate $\frac{d y}{d x}$ at $x=2$ . $\newline$ Use an exact expression.

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Find derivatives of inverse trigonometric functions

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y=\arccos \left(-\frac{x}{5}\right)

\newline

Evaluate

\frac{d y}{d x}

x=2

\newline

Use an exact expression.

Find Derivative of $y=\arccos\left(-\frac{x}{5}\right):</b> First, let's find the derivative of \$y=\arccos\left(-\frac{x}{5}\right)$ . Using the chain rule, the derivative of $\arccos(u)$ is $-\frac{1}{\sqrt{1-u^2}}$ times the derivative of $u$ .
Apply Chain Rule: Let $u=-\frac{x}{5}$ . Then, $\frac{du}{dx} = -\frac{1}{5}$ .
Plug in $u$ and Simplify: Now, plug $u$ into the derivative of $\arccos(u)$ to get $\frac{dy}{dx} = -\frac{1}{\sqrt{1-\left(-\frac{x}{5}\right)^2}} \cdot \left(-\frac{1}{5}\right)$ .
Evaluate at $x=2$ : Simplify the expression: $\frac{dy}{dx} = -\frac{1}{\sqrt{1-\left(\frac{x^2}{25}\right)}} \cdot \left(-\frac{1}{5}\right)$ .
Calculate Inside Square Root: Now, evaluate $\frac{dy}{dx}$ at $x=2$ . Plug $x=2$ into the derivative: $\frac{dy}{dx} = -\frac{1}{\sqrt{1-\left(\frac{2^2}{25}\right)}} \times \left(-\frac{1}{5}\right)$ .
Take Square Root: Calculate the inside of the square root: $1-(2^2)/(25) = 1-(4)/(25) = 1-4/25 = 21/25$ .
Multiply by $-\frac{1}{5}$ : Now, take the square root: $\sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$ .
Multiply by $-\frac{1}{5}$ : Now, take the square root: $\sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$ .Finally, multiply by $-\frac{1}{5}$ : $\frac{dy}{dx} = -\frac{1}{\sqrt{21}/5} \times (-\frac{1}{5}) = \frac{1}{5\sqrt{21}/5} = \frac{1}{\sqrt{21}}$ .

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y=arccos⁡(−x5) y=\arccos \left(-\frac{x}{5}\right) y=arccos(−5x​)\newlineEvaluate dydx \frac{d y}{d x} dxdy​ at x=2 x=2 x=2.\newlineUse an exact expression.

$y=\arccos \left(-\frac{x}{5}\right)$ $\newline$ Evaluate $\frac{d y}{d x}$ at $x=2$ . $\newline$ Use an exact expression.