1. ## Inverse Cosine Derivative

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

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

We know this rule:

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

So.....

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

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

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

What's going on here?

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

Multipy that by

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

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

Multipy that by

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

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

4. I think so

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

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

Yup!

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

--Chris