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.
Hello, KarlosK!
Exactly where is your difficulty?
. . Log rules?
. . The derivative of logs?
. . You really suck at algebra?
Simplify first . . .Differentiate: .$\displaystyle y \:=\:\ln\left(\frac{1}{x\sqrt{x + 1}}\right)$
The answer in the back of the book has: $\displaystyle \frac{3x + 2}{2x(x+1)}$
. . This should have a "minus" in front.
$\displaystyle y \;=\;\ln\left(\frac{1}{x\sqrt{x+1}}\right) \;=\;\underbrace{\ln(1)}_{\text{This is 0}} - \ln\left(x\sqrt{x+1}\right) \;=\;-\ln\left(x\sqrt{x+1}\right)$
. $\displaystyle =\;-\ln\left[x(x+1)^{\frac{1}{2}}\right] \;=\; -\bigg[\ln x + \ln(x+1)^{\frac{1}{2}}\bigg]$
Now differentiate:
. . $\displaystyle 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)} $