Question

If $y=\log \left[\frac{x+\sqrt{x^2+25}}{\sqrt{x^2+25-x}}\right]$ , find $\frac{dy}{dx}$

Answer 1

$y=\log \left[\frac{x+\sqrt{x^2+25}}{\sqrt{x^2+25}-x}\right]$

$e^y=\left[\frac{x+\sqrt{x^2+25}}{\sqrt{x^2+25}-x}\right]$

$\Rightarrow \frac{e^y+1}{e^y-1}=\frac{2\sqrt{x^2+25}}{2x}=\sqrt{1+\frac{25}{x^2}}$

$\Rightarrow \left[\frac{e^y(e^y-1)-e^y(e^y+1)}{(e^y-1{)}^2}\right]\frac{dy}{dx}=\frac{\frac{-50}{x^3}}{2\sqrt{1+\frac{25}{x^2}}}$

$\Rightarrow \frac{-2e^y}{(e^y-1{)}^2}\frac{dy}{dx}=\frac{\frac{-25}{x^3}}{\frac{\sqrt{x^2+25}}{x}}$

$\Rightarrow \frac{dy}{dx}=\frac{25}{x^2\sqrt{x^2+25}}.\frac{(e^y-1{)}^2}{2e^y}$

$\Rightarrow =\frac{dy}{dx}=\frac{25}{x^2\sqrt{x^2+25}}=\frac{(\frac{x+\sqrt{x^2+25}}{\sqrt{x^2+25-x}}-1{)}^2}{2.\frac{x+\sqrt{x^2+25}}{\sqrt{x^2+25}-x}}$

$\Rightarrow \frac{dy}{dx}=\frac{25(x^2+25)-x}{2x^2\sqrt{x^2+25}(x+\sqrt{x^2+25})}.(\frac{2x}{x^2+25}{)}^2$

$=\frac{25\times 2}{\left(\sqrt{x^2+25}\right)(x+\sqrt{x^2+25})(\sqrt{x^2+25}-x)}$

$=\frac{25\times 2}{\sqrt{x^2+25(x^2+25-x^2)}}$

$=\frac{2}{\sqrt{x^2+25}}$