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Brush up on simplification

  • Thread starter mathor345
  • Start date
  • #1
16
0
This is a question on the simplification operations. I can't for the life of me figure out how:

[tex]\frac{1}{\frac{1}{2t^3}\sqrt{(\frac{1}{2t^3})^2 -1}}*(-\frac{3}{2t^4})= -\frac{3}{t\sqrt{\frac{1}{4t^6}(1 - 4t^6)}}[/tex]

Really, I can't figure out where [tex](1-4t^6)[/tex] is coming from!

It's involved in finding the derivative of inverse trigonometric function and I'm getting stuck right in the middle with the easy stuff.
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
5
[tex] \left( \frac{1}{2t^3} \right)^2 -1 = \frac{1-4t^6}{4t^6} [/tex] just by pulling a factor of [itex] \frac{1}{4t^6} [/itex] out to the front.
 
  • #3
eumyang
Homework Helper
1,347
10
Well, note that
[tex]\left(\frac{1}{2t^3} \right)^2 = \frac{1}{4t^6}[/tex]

So, looking inside that square root in the denominator...
[tex]\begin{aligned}
\left(\frac{1}{2t^3} \right)^2 - 1 &= \frac{1}{4t^6} - 1 \\
&= \frac{1}{4t^6} - \frac{4t^6}{4t^6} \\
&= \frac{1}{4t^6}(1) - \frac{1}{4t^6}(4t^6) \\
&= \frac{1}{4t^6}(1 - 4t^6)
\end{aligned}[/tex]

EDIT: Beaten to it. :wink:
 
  • #4
16
0
Doh! Thanks guys :P
 

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