# Thread: How to do this Inverse Trig Function?

1. ## How to do this Inverse Trig Function?

Find the derivative of y
y=sin^-1 3/t^2
All I know is the derivative of sin^-1 is 1/square root of 1-x^2

2. You know that: $\frac{d}{dx} \sin^{-1} ({\color{red}u}) = \frac{1}{\sqrt{1-{\color{red}u}^2}} \cdot ({\color{red}u})'$

Making the analogy: $y' = \frac{1}{\sqrt{1 - \left({\color{red}\frac{3}{t^2}}\right)^2}} \cdot \left({\color{red}\frac{3}{t^2}}\right)'$

3. Originally Posted by JimDavid
All I know is the derivative of sin^-1 is 1/square root of 1-x^2
Im assuming you are asking to find $\frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]$?

You have two options...you could use the fact that $\arcsin\left(\frac{1}{x}\right)=\csc^{-1}\left(x\right)$ to eliminate the problem negative exponent...this is assuming you know that derivative...or just use the chain rule

$\frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]=\frac{1}{\sqrt{1-\left(\frac{3}{x^2}\right)^2}}\cdot\left[\frac{3}{x^2}\right]'$

EDIT: The war begins anew...

4. Originally Posted by Mathstud28
Im assuming you are asking to find $\frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]$?

You have two options...you could use the fact that $\arcsin\left(\frac{1}{x}\right)=\csc^{-1}\left(x\right)$ to eliminate the problem negative exponent...this is assuming you know that derivative...or just use the chain rule

$\frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]=\frac{1}{\sqrt{1-\left(\frac{3}{x^2}\right)^2}}\cdot\left[\frac{3}{x^2}\right]'$

EDIT: The war begins anew...
I haven't learn about arcsin yet, just the basic 6 tri functions.

5. Originally Posted by JimDavid
I haven't learn about arcsin yet, just the basic 6 tri functions.
$\arcsin(x)=sin^{-1}(x)$