If we take the derivative of a typical inverse function we will end up with one divided by the square root of one minus x squared. As for example let us use the inverse of simply the sin x. Here is an example which I can easily follow. I will provide a link, it is much better than my graphics can do:
http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx
You will notice that it invokes the Inverse Function Theorem stated in the beginning which I believe the Implicit Function Theorem can be used to proof this.
I am wondering if differentiating a typical inverse trig function can be done without invoking the IFT. I suspect not but the proof of IFT is not usually given in typical first semester multivariable calculus classes even though differentiating inverse trig function is standard fair. Just curious.
You can do this with the chain rule for the function of a function ie $$\frac d{dx}g(f(x))=\frac {df(x)}{dx}\frac {d g(f(x))}{df(x)}$$ and this is normally proved (though some care is required over conditions) alongside the other elementary results like the derivative of a product.
The way it works is this, for example: suppose $y=\arcsin x$ so that $x=\sin y$. We differentiate both sides of this second equation with respect to $x$ to obtain $$1=\frac {dy}{dx}\cos y=\frac {dy}{dx}\sqrt {1-x^2}$$ so that $$\frac {dy}{dx}=\frac 1{\sqrt {1-x^2}}$$Again care is required over conditions and things like the sign of the square root.