# Derivatives of Inv Trig Functions

• Mar 18th 2008, 08:57 PM
kid funky fried
Derivatives of Inv Trig Functions
What am I doing wrong in this problem?
• Mar 18th 2008, 09:05 PM
TheEmptySet
Quote:

Originally Posted by kid funky fried
What am I doing wrong in this problem?

Try this...

$y=\sin^{-1}(x/2) \iff \sin(y)=\frac{x}{2}$ by implicit

$\cos(y) \cdot \frac{dy}{dx}=\frac{1}{2}$ solve for the derivative

$\frac{dy}{dx}=\frac{1}{2 \cos(y)}$

note that $\cos(y) =\sqrt{\cos^2(y)}=\sqrt{1-sin^2(y)}=\sqrt{1-(\frac{x}{2})^2}=\frac{1}{2}\sqrt{4-x^2}$

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

I hope this helps.
• Mar 18th 2008, 09:07 PM
o_O
$y = arcsin^{-1}\left(\frac{x}{2}\right)$
$y' = \frac{1}{\sqrt{1 - \left(\frac{x}{2}\right)^{2}}} \cdot \frac{d}{dx} \left(\frac{x}{2}\right)$
$y' = \frac{1}{\sqrt{1 - \frac{x^{2}}{4}}} \cdot \frac{1}{2}$
$y' = \frac{1}{2 \sqrt{\frac{1}{4}(4 - x^{2})}}$
$y' = \frac{1}{2 \cdot \frac{1}{2}\sqrt{4 - x^{2}}}$ You factor out a 1/2, not a 2 in the denominator
$y' = \frac{1}{\sqrt{4 - x^{2}}}$
• Mar 18th 2008, 09:15 PM
kid funky fried
ha!
Great, I see now!
Thanks alot!