# Thread: Derivatives of Inv Trig Functions

1. ## Derivatives of Inv Trig Functions

What am I doing wrong in this problem?

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

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

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

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

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

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

I hope this helps.

3. $\displaystyle y = arcsin^{-1}\left(\frac{x}{2}\right)$
$\displaystyle y' = \frac{1}{\sqrt{1 - \left(\frac{x}{2}\right)^{2}}} \cdot \frac{d}{dx} \left(\frac{x}{2}\right)$
$\displaystyle y' = \frac{1}{\sqrt{1 - \frac{x^{2}}{4}}} \cdot \frac{1}{2}$
$\displaystyle y' = \frac{1}{2 \sqrt{\frac{1}{4}(4 - x^{2})}}$
$\displaystyle y' = \frac{1}{2 \cdot \frac{1}{2}\sqrt{4 - x^{2}}}$ You factor out a 1/2, not a 2 in the denominator
$\displaystyle y' = \frac{1}{\sqrt{4 - x^{2}}}$

4. ## ha!

Great, I see now!
Thanks alot!