Find dy/dx for the following curve:
x = ( 1 - a sq ) to the power of 3/2
y = sin to the power of -1 ( a )
$\displaystyle x=(1-a^2)^{\frac{3}{2}}$
$\displaystyle y=sin^{-1}(a)$
Substitute $\displaystyle a=sin \theta$
$\displaystyle y=\theta$
$\displaystyle x=(1-sin^2\theta)^{\frac{3}{2}}$
$\displaystyle x=cos^3\theta$
$\displaystyle \frac{dx}{d\theta}=3cos^2\theta*-sin\theta$
$\displaystyle \frac{dx}{dy}=3cos^2\theta*-sin\theta$
$\displaystyle \frac{dx}{dy}=3(1-a^2)*-a$
$\displaystyle \frac{dx}{dy}=3a(a^2-1)$
$\displaystyle \frac{dy}{dx}=\frac{1}{3a(a^2-1)}$