derivative of a logarithmic function

**Stuck Man** · January 4th 2012, 09:14 AM

How is it done?

**Stuck Man** · January 4th 2012, 09:36 AM

I have done it. It is done the usual way, derivative of (x + (x^2+k)^0.5)/(x + (x^2+k)^0.5).

**MarkFL** · January 4th 2012, 09:38 AM

Use:

$\frac{d}{dx}\left(\ln(f(x))\right)=\frac{1}{f(x)} \cdot\frac{df}{dx}$

**Soroban** · January 4th 2012, 09:52 AM

Hello, Stuck Man!

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

First, you do the Calculus (find the derivative),
. . then do some Algebra.

We have: . $y \:=\:\ln\left|x + (x^2+k)^{\frac{1}{2}}\right|$

The derivative is: . $y' \:=\:\frac{1+\frac{1}{2}(x^2+k)^{-\frac{1}{2}}\cdot2x}{x+(x^2+k)^{\frac{1}{2}}} \;=\;\dfrac{1 + \dfrac{x}{\sqrt{x^2+k}}}{x+\sqrt{x^2+k}}$

Multiply by $\frac{\sqrt{x^2+k}}{\sqrt{x^2+k}}:\;\;\frac{\sqrt{ x^2+k}}{\sqrt{x^2+k}}\cdot\frac{1 + \dfrac{x}{\sqrt{x^2+k}}}{x + \sqrt{x^2+k}}$

. . . . . . . . . . . $=\;\frac{\sqrt{x^2+k}+x}{\sqrt{x^2+k}\,(x + \sqrt{x^2+k})}$

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

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