d-dx-ln-x-2-1-x-2-1-

Question Number 68703 by Maclaurin Stickker last updated on 15/Sep/19

\frac{d}{d x} (l n (\sqrt{\frac{x^{2} - 1}{x^{2} + 1}})) = ?

Commented by MJS last updated on 15/Sep/19

\frac{d}{d x} [\ln \sqrt{\frac{x^{2} - 1}{x^{2} + 1}}] = \frac{1}{2} \times \frac{d}{d x} [\ln \frac{x^{2} - 1}{x^{2} + 1}] =

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

= \frac{2 x}{x^{4} - 1}

Commented by Prithwish sen last updated on 15/Sep/19

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

= \sqrt{\frac{x^{2} + 1}{x^{2} - 1}} . \frac{\frac{x \sqrt{x^{2} + 1}}{\sqrt{x^{2} - 1}} - \frac{x \sqrt{x^{2} - 1}}{\sqrt{x^{2} + 1}}}{x^{2} + 1}

= \sqrt{\frac{x^{2} + 1}{x^{2} - 1}} . \frac{2 x}{\sqrt{x^{2} - 1} . {(x^{2} + 1)}^{\frac{3}{2}}}

= \frac{2 x}{(x^{2} - 1) (x^{2} + 1)} = \frac{2 x}{(x^{4} - 1)} pleasecheck .

Commented by Maclaurin Stickker last updated on 15/Sep/19

Y o u r a n s w e r i s r i g h t!

Commented by Prithwish sen last updated on 15/Sep/19

Thanks Sir .

Commented by mathmax by abdo last updated on 15/Sep/19

l e t f (x) = l n (\sqrt{\frac{x^{2} - 1}{x^{2} + 1}}) \Rightarrow f (x) = \frac{1}{2} {l n (x^{2} - 1) - l n (x^{2} + 1)} \Rightarrow

f^{^{'}} (x) = \frac{1}{2} {\frac{2 x}{x^{2} - 1} - \frac{2 x}{x^{2} + 1}} = x {\frac{1}{x^{2} - 1} - \frac{1}{x^{2} + 1}}

= x {\frac{x^{2} + 1 - x^{2} + 1}{x^{4} - 1}} = \frac{2 x}{x^{4} - 1}