Question

If $y=\frac{\sqrt{(a^2+x^2)}+\sqrt{(a^2-x^2)}}{\sqrt{(a^2+x^2)}-\sqrt{(a^2-x^2)}}$, then show that
$\frac{dy}{dx}=-\frac{2a^2}{x^3}\{1+\frac{a^2}{\sqrt{(a^4-x^4)}}\}$

Answer 1

$y=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}$
Multiply numerator and denominator by $\sqrt{a^2+x^2}+\sqrt{a^2-x^2}$

$y=\frac{(\sqrt{a^2+x^2}+\sqrt{a^2-x^2}{)}^2}{a^2+x^2-(a^2-x^2)}$

$\therefore y=\frac{(a^2+x^2)+(a^2-x^2)+2\sqrt{(a^4-x^4)}}{2x^2}$

$y=\frac{a^2}{x^2}+\frac{\sqrt{(a^4-x^4)}}{x^2}$

$\therefore \frac{dy}{dx}=-\frac{2a^2}{x^3}-2\frac{\sqrt{(a^4-x^4)}}{x^3}+\frac{1}{x^2}.\frac{1}{2\sqrt{(a^4-x^4)}}(-4x^3)$

$=-\frac{2a^2}{x^3}[1+\frac{a^2}{\sqrt{(a^4-x^4)}}]$