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${:[h(x)=arctan(-(x)/(2))],[h^(')(-7)=◻]:} Use an exact expression.$

$\begin{array}{l} h(x)=\arctan \left(-\frac{x}{2}\right) \\ h^{\prime}(-7)=\square \end{array}$ $\newline$ Use an exact expression.

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Find derivatives of inverse trigonometric functions

Full solution

Q.

\begin{array}{l} h(x)=\arctan \left(-\frac{x}{2}\right) \\ h^{\prime}(-7)=\square \end{array}

\newline

Use an exact expression.

Find derivative using chain rule: step_1: Find the derivative of $h(x) = \arctan\left(-\frac{x}{2}\right)$ . To find the derivative of $h(x)$ , we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Apply chain rule to $h(x)$ : step_2: Apply the chain rule to $h(x) = \arctan\left(-\frac{x}{2}\right)$ . The outer function is $\arctan(u)$ , and the inner function is $u = -\frac{x}{2}$ . The derivative of $\arctan(u)$ with respect to $u$ is $\frac{1}{1+u^2}$ . The derivative of $u = -\frac{x}{2}$ with respect to $x$ is $-\frac{1}{2}$ .
Combine derivatives to find $h'(x)$ : step_3: Combine the derivatives to find $h'(x)$ .
$h'(x) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$
$= \frac{1}{1+\left(-\frac{x}{2}\right)^2} \cdot \left(-\frac{1}{2}\right)$
$= \frac{1}{1+\frac{x^2}{4}} \cdot \left(-\frac{1}{2}\right)$
$= -\frac{1}{2\cdot(1+\frac{x^2}{4})}$
$= -\frac{1}{2+\frac{x^2}{2}}$
Simplify expression for $h'(x)$ : step_4: Simplify the expression for $h'(x)$ . $h'(x) = -\frac{1}{2+(\frac{x^2}{2})}$ $= -\frac{1}{2(1+\frac{x^2}{4})}$ $= -\frac{1}{2+\frac{x^2}{4}}$
Evaluate $h'(x)$ at $x = -7$ : step_5: Evaluate $h'(x)$ at $x = -7$ . $h'(-7) = -1/(2+(-7)^2/4) = -1/(2+49/4) = -1/(2+12.25) = -1/(14.25) = -1/14.25$

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