# Derivative of inverse trig functions.

## Homework Statement

Find the derivative:
y=sec-1(1/2t3)

## Homework Equations

$$\frac{\frac{du}{dx}}{|u|\sqrt{u^2-1}}$$

## The Attempt at a Solution

I have an example to follow, but I dont know how step 1. became step 2.?...or more exactly the last part under the radical? (1-4t^6) instead of just -1 ???

1.$$y'=\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{(\frac{1}{2t^3})^2-1}}$$

2.$$y'=\frac{-3}{(t)\sqrt{\frac{1}{4t^6}(1-4t^6)}}$$

## Answers and Replies

Nevermind... I see it now.