1. Inverse Cosine Derivative

This is stumping me. Find the derivative of the following function:

$
y = \cos ^{ - 1} \left( {\frac{x}
{2}} \right)
$

We know this rule:

$
\frac{d}
{{dx}}\cos ^{ - 1} u = \frac{{ - du/dx}}
{{\sqrt {1 - u^2 } }},|u| < 1
$

So.....

$
\begin{gathered}
u = \tfrac{x}
{2} \hfill \\
\tfrac{{du}}
{{dx}} = \tfrac{1}
{2} \hfill \\
\frac{d}
{{dx}}\cos ^{ - 1} u = \frac{{ - 1/2}}
{{\sqrt {1 - \tfrac{{x^2 }}
{4}} }} = \frac{{ - 1/2}}
{{\sqrt {4 - x^2 } }} \hfill \\
\end{gathered}
$

But this is wrong....it should be....

$
y = \cos ^{ - 1} \left( {\frac{x}
{2}} \right) = \frac{{ - 1}}
{{\sqrt {4 - x^2 } }}
$

What's going on here?

2. $\frac{{ - 1/2}}
{{\sqrt {1 - \tfrac{{x^2 }}
{4}} }}$

Multipy that by

$\frac{\sqrt{4}}{\sqrt{4}}$

3. Originally Posted by 11rdc11
$\frac{{ - 1/2}}
{{\sqrt {1 - \tfrac{{x^2 }}
{4}} }}$

Multipy that by

$\frac{\sqrt{4}}{\sqrt{4}}$
So this is true as well? Given a as a constant.

$
\frac{d}
{{dx}}\cos ^{ - 1} \frac{x}
{a} = \frac{{ - 1}}
{{\sqrt {a^2 - x^2 } }}
$

4. I think so

5. Originally Posted by RedBarchetta
So this is true as well? Given a as a constant.

$
\frac{d}
{{dx}}\cos ^{ - 1} \frac{x}
{a} = \frac{{ - 1}}
{{\sqrt {a^2 - x^2 } }}
$

Yup!

Because $\frac{d}{\,dx}\cos^{-1}\left(\frac{x}{a}\right)=-\frac{1}{a}\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}=-\frac{1}{\sqrt{a^2}}\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}=-\frac{1}{\sqrt{a^2\left(1-\frac{x^2}{a^2}\right)}}=-\frac{1}{\sqrt{a^2-x^2}}$

--Chris