Posted by carolina on Monday, October 26, 2009 at 5:39pm.
What is under the square root sign? the way you have written it, it is the argument of the sine function, 2t
Assuming that is it
y= INVsin(sqrt2t)
I am going to shift to arcsin notation, so we dont get messed up in power notation.
y=arcsin(sqrt2t)
y=arcsin u where u=sqrt2t
y'=1/(1-u^2) du/dx
and du/dt= 1/(2sqrt2t) * 2 or 1/sqrt2t
y=1/((1-2t)(sqrt2t))
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