In another way, we know that
tan(theta) = sin(theta)/cos(theta)tan(θ)=sin(θ)cos(θ)
And that cos(theta) = sqrt(1 - sin^2(theta))cos(θ)=√1−sin2(θ)
So we can say that
tan(theta) = sin(theta)/sqrt(1-sin^2(theta))tan(θ)=sin(θ)√1−sin2(θ)
For theta = arcsin(x)θ=arcsin(x) we have
tan(arcsin(x)) = x/sqrt(1-x^2)tan(arcsin(x))=x√1−x2
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 chain rule we have
y = x/sqrt(1 - x^2)y=x√1−x2
dy/dx = 1/sqrt(1-x^2)d/dxx+xd/dx(1/sqrt(1-x^2))dydx=1√1−x2ddxx+xddx(1√1−x2)
dy/dx = 1/sqrt(1-x^2)+xd/dx(1/sqrt(1-x^2))dydx=1√1−x2+xddx(1√1−x2)
Say 1 - x^2 = u1−x2=u
dy/dx = 1/sqrt(1-x^2)+xd/(du)(1/sqrt(u))(du)/dxdydx=1√1−x2+xddu(1√u)dudx
dy/dx = 1/sqrt(1-x^2)+x(-1/(2u^(3/2)))(-2x)dydx=1√1−x2+x(−12u32)(−2x)
dy/dx = 1/sqrt(1-x^2) + (x^2)/((1-x^2)sqrt(1-x^2)dydx=1√1−x2+x2(1−x2)√1−x2
From there it's just algebra (also, as a sidenote, that (1-x^2)(1−x2) would be in absolute value bars but since it's always positive for the range of xx 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))dydx=(1−x2)+x2(1−x2)√1−x2=1(1−x2)√1−x2
Or, if you prefer
dy/dx = (1-x^2)^(-3/2)dydx=(1−x2)−32
