What is the derivative of ##y=tan(arcsin (x))##?

##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)##