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 don't 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))
can someone actually show me all the steps to finding this deriviative. I keep doing the problem out and getting the wrong answer.
Find the deriviative of:
INVERSEsin(2t)^1/2
1 answer
1 answer