##dy/dx = (1-x^2)^(-3/2)##
In another way, we know that
##tan(theta) = sin(theta)/cos(theta)##
And that ##cos(theta) = sqrt(1 – sin^2(theta))##
So we can say that
##tan(theta) = sin(theta)/sqrt(1-sin^2(theta))##
For ##theta = arcsin(x)## we have
##tan(arcsin(x)) = x/sqrt(1-x^2)##
Which is simply a rational function and won’t touch on messy trig derivates, and is easy to derive using the appropriate tricks. Using only the we have
##y = x/sqrt(1 – x^2)##
##dy/dx = 1/sqrt(1-x^2)d/dxx+xd/dx(1/sqrt(1-x^2))##
##dy/dx = 1/sqrt(1-x^2)+xd/dx(1/sqrt(1-x^2))##
Say ##1 – x^2 = u##
##dy/dx = 1/sqrt(1-x^2)+xd/(du)(1/sqrt(u))(du)/dx##
##dy/dx = 1/sqrt(1-x^2)+x(-1/(2u^(3/2)))(-2x)##
##dy/dx = 1/sqrt(1-x^2) + (x^2)/((1-x^2)sqrt(1-x^2)##
From there it’s just algebra (also, as a sidenote, that ##(1-x^2)## would be in absolute value bars but since it’s always positive for the range of ##x## we can take, we don’t bother with it.
##dy/dx = ((1-x^2)+x^2)/((1-x^2)sqrt(1-x^2)) = 1/((1-x^2)sqrt(1-x^2))##
Or, if you prefer
##dy/dx = (1-x^2)^(-3/2)##
