Yeah, I tried doing this question and just wanted to check if I was correct.(adsbygoogle = window.adsbygoogle || []).push({});

Write down the derivative of [tex]\arccos (4x^2)[/tex] and state the domain for which the derivative applies.

[tex]\mbox{let}\ y = \arccos (4x^2), x\ \epsilon\ [-1, 1], y\ \epsilon\ [0, \pi][/tex]

[tex]\Leftrightarrow x = \frac{\sqrt{\cos y}}{2}[/tex]

[tex]\begin{align*}

\frac{d}{dx} x = \frac{d}{dx} \frac{\sqrt{\cos y}}{2} \\

1 = \frac{-\sin y}{4 \sqrt{\cos y}} \cdot \frac{dy}{dx} \\

\frac{dy}{dx} = \frac{4 \sqrt{\cos y}}{-\sin y}

\end{align*}[/tex]

[tex]\cos^2 y + \sin^2 y = 1[/tex]

[tex]\sin y = \sqrt{1 - \cos^2 y}, \mbox{because}\ y\ \epsilon\ [0, \pi]\ \mbox{so}\ \sin y \geq 0[/tex]

[tex]\therefore \frac{dy}{dx} = \frac{4\sqrt{\cos y}}{-\sqrt{1 - \cos^2 y}}[/tex]

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# Homework Help: Inverse Cos Derivative

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