Question

Which equation is equivalent to \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\] for \[y=\sin^{-1}x\]?

Options

• \[(1-x^2)\frac{dy}{dx}=1\]

• \[\sqrt{1-x^2}\frac{dy}{dx}=1\]

• \[\sqrt{1-x^2}+\frac{dy}{dx}=1\]

• \[\sqrt{1-x^2}\frac{d^2y}{dx^2}=1\]

Answer 1

Multiplying \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\] by \[\sqrt{1-x^2}\] gives the required equation. The product of \[\sqrt{1-x^2}\] and its reciprocal is \[1\].