1. ## Double Check

Hi

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

${{d}\over{d\,x}}\,{\it arctan}\left({{x}\over{\sqrt{1-x^2}}}\right)$ = $\sqrt{1-x^2}$

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

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

2. Originally Posted by xwrathbringerx
Hi

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

${{d}\over{d\,x}}\,{\it arctan}\left({{x}\over{\sqrt{1-x^2}}}\right)$ = $\sqrt{1-x^2}$

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

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

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

3. Ummm what's wrong with the second answer?

4. Originally Posted by xwrathbringerx
Hi

$\frac{d}{dx}*\frac{1}{arcsinx} = \frac{-1}{(arcsin)^2*\sqrt{1-x^2}}$
arcsin(of what?) in the denominator

5. it's arcsin of x.

Sori

6. $y=tan^{'}f(x)$

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

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

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

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

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

If you're wondering how to do it.