# Double Check

• Nov 15th 2009, 04:37 PM
xwrathbringerx
Double Check
Hi

Could someone check to c if these are the right answers:

http://www.freemathhelp.com/cgi-bin/...^2}}}\right)$$= \displaystyle \sqrt{1-x^2} \displaystyle \frac{d}{dx}*\frac{1}{arcsinx} = \frac{-1}{(arcsin)^2*\sqrt{1-x^2}} If they're wrong, could someone please show me the right way to do them?? • Nov 15th 2009, 04:59 PM skeeter Quote: Originally Posted by xwrathbringerx Hi Could someone check to c if these are the right answers: http://www.freemathhelp.com/cgi-bin/...D%5Cright%29$$ = $\displaystyle \sqrt{1-x^2}$

should be $\displaystyle \textcolor{red}{\frac{1}{\sqrt{1-x^2}}}$

$\displaystyle \frac{d}{dx}*\frac{1}{\arcsin{x}} = \frac{-1}{(\arcsin{\textcolor{red}{x}})^2*\sqrt{1-x^2}}$

If they're wrong, could someone please show me the right way to do them??

...
• Nov 15th 2009, 05:02 PM
xwrathbringerx
Ummm what's wrong with the second answer?
• Nov 15th 2009, 05:04 PM
skeeter
Quote:

Originally Posted by xwrathbringerx
Hi

$\displaystyle \frac{d}{dx}*\frac{1}{arcsinx} = \frac{-1}{(arcsin)^2*\sqrt{1-x^2}}$

arcsin(of what?) in the denominator
• Nov 15th 2009, 05:07 PM
xwrathbringerx
it's arcsin of x.

Sori
• Nov 15th 2009, 05:11 PM
I-Think
$\displaystyle y=tan^{'}f(x)$

$\displaystyle tan y=f(x)$
$\displaystyle \frac{d}{dx} tan y= \frac{d}{dx} f(x)$

$\displaystyle sec^2y\frac{dy}{dx}=f^{'}(x)$
$\displaystyle (tan^2y+1)\frac{dy}{dx}=f^{'}(x)$
$\displaystyle \frac{dy}{dx}=\frac{f^{'}(x)}{[f(x)]^2+1}$

Find $\displaystyle \frac{d}{dx} arctan(\frac{x}{\sqrt{1-x^2}})$

If $\displaystyle g(x)=\frac{x}{\sqrt{1-x^2}}$
Then $\displaystyle g^{'}x=\frac{\sqrt{1-x^2}+x^2(1-x^2)^{-0.5}}{1-x^2}$
And $\displaystyle [g(x)]^2=\frac{x^2}{1-x^2}$

So
$\displaystyle \frac{d}{dx} arctan(\frac{x}{\sqrt{1-x^2}})=\frac{\frac{\sqrt{1-x^2}+x^2(1-x^2)^{-0.5}}{1-x^2}}{\frac{x^2}{1-x^2}+1}$
Which becomes
$\displaystyle \sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}$
Becoming
$\displaystyle \frac{1}{\sqrt{1-x^2}}$

If you're wondering how to do it.