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...
$$\begin{aligned} \left(\frac{1}{2t^3} \right)^2 - 1 &= \frac{1}{4t^6} - 1 \\ &= \frac{1}{4t^6} - \frac{4t^6}{4t^6} \\ &= \frac{1}{4t^6}(1) - \frac{1}{4t^6}(4t^6) \\ &= \frac{1}{4t^6}(1 - 4t^6) \end{aligned}$$

EDIT: Beaten to it.

 

[mathor345]

Doh! Thanks guys :P