# Thread: Help finding derivative of logs

1. ## Help finding derivative of logs

I have a problem that states: Find the derivative of y with respect to x

y = ln (1/x(sqroot x + 1)

The answer in the back of the book has (3x + 2) / 2x(x+1) as the answer and I don't understand how they came to this.

2. Hello, KarlosK!

. . Log rules?
. . The derivative of logs?
. . You really suck at algebra?

Differentiate: . $y \:=\:\ln\left(\frac{1}{x\sqrt{x + 1}}\right)$

The answer in the back of the book has: $\frac{3x + 2}{2x(x+1)}$
. . This should have a "minus" in front.
Simplify first . . .

$y \;=\;\ln\left(\frac{1}{x\sqrt{x+1}}\right) \;=\;\underbrace{\ln(1)}_{\text{This is 0}} - \ln\left(x\sqrt{x+1}\right) \;=\;-\ln\left(x\sqrt{x+1}\right)$

. $=\;-\ln\left[x(x+1)^{\frac{1}{2}}\right] \;=\; -\bigg[\ln x + \ln(x+1)^{\frac{1}{2}}\bigg]$

Now differentiate:

. . $y' \;=\;-\bigg[\frac{1}{x} + \frac{1}{2}\!\cdot\!\frac{1}{x+1}\bigg] \;=\;-\bigg[\frac{2(x+1) + x}{2x(x+1)}\bigg] \;=\;-\frac{3x+2}{2x(x+1)}$