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${:[g(x)=arccos(3x)],[g^(')(-(1)/(5))=◻]:} Use an exact expression.$

$\begin{array}{l} g(x)=\arccos (3 x) \\ g^{\prime}\left(-\frac{1}{5}\right)=\square \end{array}$ $\newline$ Use an exact expression.

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Find derivatives of inverse trigonometric functions

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Q.

\begin{array}{l} g(x)=\arccos (3 x) \\ g^{\prime}\left(-\frac{1}{5}\right)=\square \end{array}

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Use an exact expression.

Chain Rule Derivative Calculation: The chain rule states that if we have a composite function $g(x) = h(u(x))$ , then $g'(x) = h'(u(x)) \cdot u'(x)$ . Here, $h(x) = \arccos(x)$ and $u(x) = 3x$ . We already know that $h'(x) = -\frac{1}{\sqrt{1-x^2}}$ , so we need to differentiate $u(x) = 3x$ to get $u'(x)$ . $u'(x) = \frac{d}{dx}(3x) = 3$ . Now, we can find $g'(x)$ by multiplying $h'(u(x))$ by $u'(x)$ . $g'(x) = h'(u(x)) \cdot u'(x)$ $1$ .
Substitute $x=-\frac{1}{5}$ : To find $g'(-\frac{1}{5})$ , we substitute $x = -\frac{1}{5}$ into the derivative we found. $\newline$ $g'(-\frac{1}{5}) = -\frac{1}{\sqrt{1-(3*(-\frac{1}{5}))^2}} \times 3$ $\newline$ = $-\frac{1}{\sqrt{1-(\frac{9}{25})}} \times 3$ $\newline$ = $-\frac{1}{\sqrt{1-\frac{9}{25}}} \times 3$ $\newline$ = $-\frac{1}{\sqrt{\frac{16}{25}}} \times 3$ $\newline$ = $-\frac{1}{\frac{4}{5}} \times 3$ $\newline$ = $-1 \times \frac{5}{4} \times 3$ $\newline$ = $-\frac{15}{4}$ $\newline$ However, we made a mistake in the calculation. We should not have multiplied by $3$ again because we already included the factor of $3$ in the derivative formula. Let's correct this. $\newline$ $g'(-\frac{1}{5})$ $0$ $\newline$ = $g'(-\frac{1}{5})$ $1$ $\newline$ = $g'(-\frac{1}{5})$ $2$ $\newline$ = $g'(-\frac{1}{5})$ $3$ $\newline$ = $g'(-\frac{1}{5})$ $4$

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