Question

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

Answer 1

$\frac{d}{dx}\left(log\left(\frac{2x+2x}{\sqrt{25-x+x^2}}\right)\right)$

$=\frac{d}{dx}\left[log\right(25+2x)+log\left(\frac{1}{\sqrt{25-x+x^2}}\right)]$

$=\frac{d}{dx}\left(log\right(25+2x)+\frac{d}{dx}(log\left(\right(25-x+x^2{)}^{-1/2})$

Using chain rule

$=\frac{d}{dx}\left(log\right((25-x+x{)}^{-1/2})+\frac{\frac{d}{dx}(25+2x)}{2x+25}$

$=\frac{2}{2x+25}+\frac{d}{dx}\left(log\right((25-x+x^2{)}^{-1/2}))$

$=\frac{2}{2x+25}+\frac{d}{dx}\left(\frac{-1}{2}log\right(25-x+x^2\left)\right)$

$=\frac{2}{2x+25}+\frac{1}{2}\left(\frac{-1+2x}{25-x+x^2}\right)$

$=\frac{2}{2x+25}+\frac{1}{2}\left[\frac{2x-1}{x^2-x+25}\right]$

$=\frac{125-52x}{1250+50x+46x^2+4x^3}$.