Inverse Trig Function Problem

[Jason76]

Find the Derivative

\(\displaystyle y = 2 \arcsin \sqrt{1 - 2x}\)

using the formula for derivative of arcsin:

\(\displaystyle \frac{d}{dx}\sin^{-1} \)

and

\(\displaystyle u = (\frac{1}{\sqrt{1 - u^{2}}})(\frac{du}{dx})\)

Step 1.

\(\displaystyle y = 2[\frac{1}{\sqrt{1-(\sqrt{1-2x})^{2}}}] [\frac{1}{2}(1-2x)^{-1/2}](-2)\)

Step 2:

Can't understand how they get from here (below this sentence) to the answer.

\(\displaystyle y = (\frac{1}{\sqrt{1-1+2x}})(\frac{1}{\sqrt{1-2x}})(-2)\)

Answer:

\(\displaystyle y = \frac{-2}{\sqrt{2x - 4x^{2}}}\)

 

[Jason76]

Figured out the answer:

In the 2nd to last line (line above the Final Answer), you multiply denominator square roots. So on the left side one, 1 and -1 subracted = 0. So your left with 2x on the left side. This 2x is multiplied by the denominator numbers on the center parenthesis. Of course, as far as numerators go, you multiply 1, 1 and -2 which yields -2 for the numerator.