Question

Derivatives.
y=(sec^-1 x) × √(x^2-1)

Answer 1

since you have two functions multiplying, this will be a product rule and a chain rule for the second function:
\[\frac{d}{dx} \left [ arcsec(x) \sqrt{x^2-1} \right ] = \sqrt{x^2-1} \frac{d}{dx}\left [ arcsec(x) \right ] + arcsec(x) \frac{d}{dx} \left [ \sqrt{x^2-1} \right ]. \]

Do you have an idea where to continue from there?

Answer 2

yes, tnx, I was just making sure. That's what I have thought.

when I do the product rule though, am I able to re-write sec^-1 x
as


x
---------
sqrt(x^2-1)


being that:


which would yield that\[y'=(\frac{x}{\sqrt{x^2-1}})(\frac{d}{dx}~\sqrt{x^2-1}~)~+~(\sqrt{x^2-1}~)(\frac{d}{dx}~\sec^{-1}x~) \]
\[y'=(\frac{x}{\sqrt{x^2-1}})(\frac{1}{\sqrt{x^2-1}})~+~(\sqrt{x^2-1}~)(~\frac{x^2-1}{x})\]