Understanding Simplification Operations for Inverse Trigonometric Functions

[mathor345]

This is a question on the simplification operations. I can't for the life of me figure out how:

\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)}}

Really, I can't figure out where (1-4t^6) 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.

 

[Gib Z]

\left( \frac{1}{2t^3} \right)^2 -1 = \frac{1-4t^6}{4t^6} just by pulling a factor of \frac{1}{4t^6} out to the front.

 

[eumyang]

Well, note that
\left(\frac{1}{2t^3} \right)^2 = \frac{1}{4t^6}

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

EDIT: Beaten to it.

 

[mathor345]

Doh! Thanks guys :P