inverse trig funtion derivatives

[tsh44]

Hi could you please help me work out this problem.

find the derivative of y with respect to the appropriate variable

y = sin inverse SQRT(2t)

I reached this

1/ (SQRT ( 1-(SQRT2t)^2)) and I know I have to multiply this by the derivative of SQRT(2t) which I got 2t^-1/2 for. I am having trouble simplifying this expression from there on. thanks for any help!

 

[skeeter]

\(\displaystyle \L \frac{d}{dt} \arcsin{u} = \frac{\frac{du}{dt}}{\sqrt{1-u^2}}\)

for \(\displaystyle \L y = \arcsin{\sqrt{2t}}\)

\(\displaystyle \L u = \sqrt{2t}\) and \(\displaystyle \L \frac{du}{dt} = \frac{1}{\sqrt{2t}}\)

so ...

\(\displaystyle \L \frac{dy}{dt} = \frac{\frac{1}{\sqrt{2t}}}{\sqrt{1 - 2t}} = \frac{1}{\sqrt{2t(1-2t)}}\)