Question

If $\sqrt{1+x^2}+\sqrt{1+y^2}=a$, then $\frac{dy}{dx}=$

• A. $\sqrt{\frac{1-y^2}{1-x^2}}$
• B. $-\frac{x}{y}\sqrt{\frac{1+y^2}{1+x^2}}$
• C. $\sqrt{\frac{x^2-1}{y^2-1}}$
• D. None of these

Answer 1

The correct option is B $-\frac{x}{y}\sqrt{\frac{1+y^2}{1+x^2}}$

Given, $\sqrt{1+x^2}+\sqrt{1+y^2}=a$ where $a$ is a constant.

Differentiating both sides w.r.t $x$,

$\Rightarrow \frac{2x}{2\sqrt{1+x^2}}+\frac{2y\frac{dy}{dx}}{2\sqrt{1+y^2}}=0$

$\Rightarrow \frac{dy}{dx}=\frac{-x\sqrt{1+y^2}}{y\sqrt{1+x^2}}$