Derivatives of Inverse Trig Functions

[crayzeerunner]

The problem that I have which may seem simple is

Cos(arcsin2x)

The instructions say to Write the problem in Algebriac Form.

The first thing that I do is find what U = which is 2x.
Then I write: Cos * 1/sqrt(1-4x^2) * 2
Then I write: (Cos*(2))/sqrt(1-4x^2)

This is where i get stuck. I do not know what to do next. According to the back of my text book then answer is sqrt(1-4x^2). I do not understand where the Cos(2) went or the fraction for that matter. If anyone can help me it would be greatly appreciated. Thanks!

 

[pka]

Cos(arcsin2x)
The instructions say to Write the problem in Algebriac Form.
According to the back of my text book then answer is sqrt(1-4x^2).

If that is the exact statement of the problem in red, then the textbook’s answer is correct. There is no derivative mentioned in the instructions.
\(\displaystyle \L
\begin{array}{l}
u = \arcsin (2x)\quad \Rightarrow \quad \sin (u) = 2x = \frac{{2x}}{1} \\
\cos (u) = \frac{{\sqrt {1^2 - \left( {2x} \right)^2 } }}{1} = \sqrt {1 - 4x^2 } \\
\cos (\arcsin (2x)) = \sqrt {1 - 4x^2 } \\
\end{array}\)