## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dy}{dx}=\frac{1}{\sqrt{(1-x^2)}}$
Given $x=\sin{y}$, differentiate the equation with respect to x: $1=\cos{y}\frac{dy}{dx}$ $\frac{dy}{dx}=\frac{1}{\cos{y}}$ $\cos{y}=\sqrt{(1-\sin^2{y})}=\sqrt{(1-x^2)}$ $\frac{dy}{dx}=\frac{1}{\sqrt{(1-x^2)}}$