## Calculus (3rd Edition)

Given function is $y=\sqrt {12-4x^{2}}$ Differentiating both sides of the equation with respect to x,we have $\frac{dy}{dx}=\frac{1}{2\sqrt {12-4x^{2}}}\times-8x=\frac{-4x}{\sqrt {12-4x^{2}}}$ Substituting the value of $\frac{dy}{dx}$ and y in the given differential equation, we get $L.H.S= \sqrt {12-4x^{2}}\times\frac{-4x}{\sqrt {12-4x^{2}}}+4x=0=R.H.S$ Therefore, the given function is a solution of the given differential equation.