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

$$y'=\frac{(2x-\sin^{-1}y)\sqrt{1-y^2}}{x}$$
$$x\sin^{-1}y=1+x^2$$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$x'\sin^{-1}y+x(\sin^{-1}y)'=0+2x$$ $$\sin^{-1}y+x\times\frac{1}{\sqrt{1-y^2}}\times y'=2x$$ $$\sin^{-1}y+\frac{x}{\sqrt{1-y^2}}\times y'=2x$$ Next, we need to separate elements with $y'$ and those without $y'$ into 2 sides of the equation: $$\frac{x}{\sqrt{1-y^2}}\times y'=2x-\sin^{-1}y$$ Finally, calculate for $y'$: $$y'=\frac{2x-\sin^{-1}y}{\frac{x}{\sqrt{1-y^2}}}$$ $$y'=\frac{(2x-\sin^{-1}y)\sqrt{1-y^2}}{x}$$