# Thread: Derivative of Inverse Trig Function

1. ## Derivative of Inverse Trig Function

Should be relatively simple but not sure what to do after a few attempts...

$\displaystyle q(x)=csc^{ - 1}(x^{ - 1})$

2. Originally Posted by drahcirnaw
Should be relatively simple but not sure what to do after a few attempts...

$\displaystyle q(x)=csc^{ - 1}(x^{ - 1})$

I believe $\displaystyle \csc^{ - 1}(x^{ - 1}) = \sin^{-1}(x)$ ...

3. Ah, makes sense now that the answer is $\displaystyle \frac{1}{\sqrt{1-x^2}}$.
Could you show me the steps required to get from the original function to $\displaystyle sin^{-1}(x)$ though? Sorry for the troubles but I'm not seeing it clearly.

4. Originally Posted by drahcirnaw
Ah, makes sense now that the answer is $\displaystyle \frac{1}{\sqrt{1-x^2}}$.
Could you show me the steps required to get from the original function to $\displaystyle sin^{-1}(x)$ though? Sorry for the troubles but I'm not seeing it clearly.
not much to see, other than knowing that sine and cosecant are reciprocals ...

if $\displaystyle x = \sin{\theta}$ , then $\displaystyle \frac{1}{x} = \csc{\theta}$

inverses ...

$\displaystyle \sin^{-1}(x) = \theta$ ... $\displaystyle \csc^{-1}\left(\frac{1}{x}\right) = \theta$