Question

After differentiating \[\sqrt{1-x^2}\cdot\frac{dy}{dx}=1\], which equation results?

विकल्प

• \[\sqrt{1-x^2}\frac{d^2y}{dx^2}+\frac{dy}{dx}=1\]

• \[\sqrt{1-x^2}\frac{dy}{dx}=0\]

• \[(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=1\]

• \[\sqrt{1-x^2}\frac{d^2y}{dx^2}+\frac{dy}{dx}\frac{d}{dx}\left(\sqrt{1-x^2}\right)=0\]

Answer 1

The left-hand side is differentiated using the product rule, while the derivative of \[1\] is \[0\]. This produces a term containing \[\frac{d^2y}{dx^2}\] and a term containing \[\frac{d}{dx}(\sqrt{1-x^2})\].