Question

If $y=\sqrt{(1-x)(1+x)}$,then

• A. $(1-x^2)\frac{dy}{dx}-xy=0$
• B. $(1-x^2)\frac{dy}{dx}+xy=0$
• C. $(1-x^2)\frac{dy}{dx}-2xy=0$
• D. $(1-x^2)\frac{dy}{dx}+2xy=0$

Answer 1

The correct option is B

$(1-x^2)\frac{dy}{dx}+xy=0$

Explanation for the correct option

Here,

$\begin{array}{rcl}y& =& \sqrt{(1-\mathrm{x})(1+\mathrm{x})}\\ & =& \sqrt{1-{\mathrm{x}}^2}\end{array}$

Differentiate with respect to $x$.

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{1}{2\sqrt{1-{\mathrm{x}}^2}}·\left(-2\mathrm{x}\right)\\ & =& \frac{-\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\end{array}$

Now,

$\begin{array}{rcl}(1-{\mathrm{x}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{xy}& =& (1-{\mathrm{x}}^2)·\frac{-\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}+\mathrm{x}\sqrt{1-{\mathrm{x}}^2}\\ & =& \sqrt{1-{\mathrm{x}}^2}\left(-\mathrm{x}\right)+\mathrm{x}\sqrt{1-{\mathrm{x}}^2}\\ & =& 0\end{array}$

Hence, B is the correct option.