# derivative of a logarithmic function

• Jan 4th 2012, 09:14 AM
Stuck Man
derivative of a logarithmic function
How is it done?
• Jan 4th 2012, 09:36 AM
Stuck Man
Re: derivative of a logarithmic function
I have done it. It is done the usual way, derivative of (x + (x^2+k)^0.5)/(x + (x^2+k)^0.5).
• Jan 4th 2012, 09:38 AM
MarkFL
Re: derivative of a logarithmic function
Use:

$\displaystyle \frac{d}{dx}\left(\ln(f(x))\right)=\frac{1}{f(x)} \cdot\frac{df}{dx}$
• Jan 4th 2012, 09:52 AM
Soroban
Re: derivative of a logarithmic function
Hello, Stuck Man!

Quote:

$\displaystyle \text{Show that the derivative of }\,y \:=\:\ln|x+\sqrt{x^2+k}| \,\text{ is: }\<y' \:=\:\frac{1}{\sqrt{x^2+k}}$

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: .$\displaystyle \frac{1}{\sqrt{x^2+k}}$