How is it done?
Hello, Stuck Man!
$\displaystyle \text{Show that the derivative of }\,y \:=\:\ln|x+\sqrt{x^2+k}| \,\text{ is: }\<y' \:=\:\frac{1}{\sqrt{x^2+k}} $
First, you do the Calculus (find the derivative),
. . then do some Algebra.
We have: .$\displaystyle y \:=\:\ln\left|x + (x^2+k)^{\frac{1}{2}}\right|$
The derivative is: .$\displaystyle y' \:=\:\frac{1+\frac{1}{2}(x^2+k)^{-\frac{1}{2}}\cdot2x}{x+(x^2+k)^{\frac{1}{2}}} \;=\;\dfrac{1 + \dfrac{x}{\sqrt{x^2+k}}}{x+\sqrt{x^2+k}} $
Multiply by $\displaystyle \frac{\sqrt{x^2+k}}{\sqrt{x^2+k}}:\;\;\frac{\sqrt{ x^2+k}}{\sqrt{x^2+k}}\cdot\frac{1 + \dfrac{x}{\sqrt{x^2+k}}}{x + \sqrt{x^2+k}} $
. . . . . . . . . . . $\displaystyle =\;\frac{\sqrt{x^2+k}+x}{\sqrt{x^2+k}\,(x + \sqrt{x^2+k})} $
. . And cancel: .$\displaystyle \frac{1}{\sqrt{x^2+k}}$