# Thread: Prove formula for derivative inverse trig function

1. ## Prove formula for derivative inverse trig function

I need to prove the following. I do not know how to do that. Thanks for any help.

$
\frac{d}{dx} \left[\sin^-1\left(u \left( x\right)\right)\right] = \frac{1}{\sqrt[2]{1- \left[ u \left(x\right)\right]^2}} \frac{d}{dx} u\left(x\right)
$

2. Originally Posted by DBA
I need to prove the following. I do not know how to do that. Thanks for any help.

$
\frac{d}{dx} \left[\sin^-1\left(u \left( x\right)\right)\right] = \frac{1}{\sqrt[2]{1- \left[ u \left(x\right)\right]^2}} \frac{d}{dx} u\left(x\right)
$
let $y = \sin^{-1}(u)$ , rewrite as the inverse function $\sin{y} = u$

$\frac{d}{dx}\left[\sin{y} = u\right]$

$\cos{y} \cdot \frac{dy}{dx} = \frac{du}{dx}$

since $\sin{y} = u$ , $\cos{y} = \sqrt{1-u^2}$

$\frac{dy}{dx} = \frac{1}{\cos{y}} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$

3. Originally Posted by DBA
I need to prove the following. I do not know how to do that. Thanks for any help.

$
\frac{d}{dx} \left[\sin^-1\left(u \left( x\right)\right)\right] = \frac{1}{\sqrt[2]{1- \left[ u \left(x\right)\right]^2}} \frac{d}{dx} u\left(x\right)
$
Since you have a composition of functions, you would use the chain rule.

$y = \arcsin{(u(x))}$.

$\frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$.

Clearly, $\frac{du}{dx} = \frac{d}{dx}\left[u(x)\right]$.

To find $\frac{dy}{du}$, we have

$y = \arcsin{u}$

$\sin{y} = u$

$\frac{d}{du}(\sin{y}) = \frac{d}{du}(u)$

$\frac{dy}{du}\,\frac{d}{dy}(\sin{y}) = 1$

$\cos{y}\,\frac{dy}{du} = 1$

$\frac{dy}{du} = \frac{1}{\cos{y}}$

$\frac{dy}{du} = \frac{1}{\cos{\arcsin{u}}}$.

Now, from the Pythagorean Identity, we have

$\sin^2{x} + \cos^2{x} = 1$

$\cos{x} = \sqrt{1 - \sin^2{x}}$

So $\frac{dy}{du} = \frac{1}{\sqrt{1 - (\sin{\arcsin{u}})^2}}$

$= \frac{1}{\sqrt{1 - u^2}}$.

Therefore:

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - [u(x)]^2}}\,\frac{d}{dx}[u(x)]$.

4. Originally Posted by skeeter
let $y = \sin^{-1}(u)$ , rewrite as the inverse function $\sin{y} = u$

$\frac{d}{dx}\left[\sin{y} = u\right]$

$\cos{y} \cdot \frac{dy}{dx} = \frac{du}{dx}$

since $\sin{y} = u$ , $\cos{y} = \sqrt{1-u^2}$

$\frac{dy}{dx} = \frac{1}{\cos{y}} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
How do I know that $\cos{y} = \sqrt{1-u^2}$

And I get confused with the u(x). Now we just use u. How do I end up with u(x) again. I do not get that at all.
Would you mind explaining that a little more?
Thanks

5. Originally Posted by DBA
How do I know that $\cos{y} = \sqrt{1-u^2}$

because $\textcolor{red}{\cos^2{y} + \sin^2{y} = 1}$

And I get confused with the u(x). Now we just use u. How do I end up with u(x) again. I do not get that at all.
Would you mind explaining that a little more?

u(x) just means that u is a function of x ... writing u by itself means the same thing.
...

6. Originally Posted by skeeter
...
Oh, ok. I got it now. Thank you!