Derivative of inverse trig functions.

  • Thread starter jrjack
  • Start date
  • #1
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Homework Statement


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

Homework Equations



[tex]\frac{\frac{du}{dx}}{|u|\sqrt{u^2-1}}[/tex]

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.[tex]y'=\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{(\frac{1}{2t^3})^2-1}}[/tex]

2.[tex]y'=\frac{-3}{(t)\sqrt{\frac{1}{4t^6}(1-4t^6)}}[/tex]
 

Answers and Replies

  • #2
111
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Nevermind... I see it now.
 

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