Question

Differentiate $\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ w.r.t. $\sqrt{1-x^4}$

Answer 1

Let $y=\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$

squaring on both sides

$y^2=\frac{(\sqrt{1+x^2}+\sqrt{1-x^2}{)}^2}{(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^2}=\frac{1+x^2+1-x^2+2\sqrt{1-x^4}}{1+x^2+1-x^2-2\sqrt{1-x^4}}=\frac{1+\sqrt{1-x^4}}{1-\sqrt{1-x^4}}$

Let $t=\sqrt{1-x^4}$

$⟹y^2=\frac{1+t}{1-t}$

Differentiating on both sides w.r.t t

$2y\frac{dy}{dt}=\frac{1-t+1+t}{(1-t{)}^2}=\frac{2}{(1-t{)}^2}$

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

$=\frac{1}{(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^{\frac{1}{2}}(1+x^2-1+x^2)}$

$=\frac{1}{2x^2(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^{\frac{1}{2}}}$