Derivative using Logarithmic differentation

[TommG]

Need to find derivative using logarithmic differentiation
$y = \sqrt{x(x+1)}$

My attempt
$ln y = ln \sqrt{x(x+1)}$

$ln y = \frac{1}{2}ln x(x+1)$

$ln y = \frac{1}{2}ln x + ln(x+1)$

$\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}$

$\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}$

$\frac{dy}{dx}= y(\frac{1}{2x} + \frac{1}{x+1})$

$\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{1}{2x} + \frac{1}{x+1})$

$\frac{dy}{dx}= \frac{\sqrt{x(x+1)}}{2x} + \frac{\sqrt{x(x+1)}}{x+1}$

answer in book $\frac{2x+1}{2\sqrt{x(x+1))}}$

 

[verty]

Need to find derivative using logarithmic differentiation
$y = \sqrt{x(x+1)}$

My attempt
$ln y = ln \sqrt{x(x+1)}$

$ln y = \frac{1}{2}ln x(x+1)$

$ln y = \frac{1}{2}ln x + ln(x+1)$

$\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}$

$\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}$

$\frac{dy}{dx}= y(\frac{1}{2x} + \frac{1}{x+1})$

$\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{1}{2x} + \frac{1}{x+1})$

$\frac{dy}{dx}= \frac{\sqrt{x(x+1)}}{2x} + \frac{\sqrt{x(x+1)}}{x+1}$

answer in book $\frac{2x+1}{2\sqrt{x(x+1))}}$

You skipped a few steps, do it more carefully and you'll see that there were mistakes.

 

[TommG]

You skipped a few steps, do it more carefully and you'll see that there were mistakes.

I made some corrections

$ln y = ln \sqrt{x(x+1)}$

$ln y = \frac{1}{2}ln x(x+1)$

$ln y = \frac{1}{2}ln x + ln(x+1)$

$\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}$

$\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}$

$\frac{1}{y}= \frac{1}{x} + \frac{1}{x+1}$

$\frac{dy}{dx}= y(\frac{1}{x} + \frac{1}{x+1})$

$\frac{dy}{dx}= y(\frac{x+1+x}{x(x+1)})$

$\frac{dy}{dx}= y(\frac{2x+1}{x(x+1)})$

$\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{2x+1}{x(x+1)})$

$\frac{dy}{dx}= (\frac{\sqrt{x(x+1)}(2x+1)}{x(x+1)})$

answer in book $\frac{2x+1}{2\sqrt{x(x+1))}}$

 

[verty]

Can you see that your answer is twice as large as the book's answer? You've lost a factor of 1/2 somewhere.

 

[Matterwave]

$$a\ln(bc)=a\ln(b)+a\ln(c)\neq a\ln(b)+\ln(c)$$

Somewhere in your derivation you made this mistake. Hopefully you can find it.

 

[HallsofIvy]

I made some corrections

$ln y = ln \sqrt{x(x+1)}$

$ln y = \frac{1}{2}ln x(x+1)$

$ln y = \frac{1}{2}ln x + ln(x+1)$

This is where you made your mistake.
$ln(y)= \frac{1}{2}(ln(x)+ ln(x+ 1))= \frac{1}{2}ln(x)+ \frac{1}{2}ln(x+ 1)$

$\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}$

$\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}$

$\frac{1}{y}= \frac{1}{x} + \frac{1}{x+1}$

$\frac{dy}{dx}= y(\frac{1}{x} + \frac{1}{x+1})$

$\frac{dy}{dx}= y(\frac{x+1+x}{x(x+1)})$

$\frac{dy}{dx}= y(\frac{2x+1}{x(x+1)})$

$\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{2x+1}{x(x+1)})$

$\frac{dy}{dx}= (\frac{\sqrt{x(x+1)}(2x+1)}{x(x+1)})$

answer in book $\frac{2x+1}{2\sqrt{x(x+1))}}$

 

[Quesadilla]

Need to find derivative using logarithmic differentiation
$y = \sqrt{x(x+1)}$

My attempt
$ln y = ln \sqrt{x(x+1)}$

$ln y = \frac{1}{2}ln x(x+1)$

$ln y = \frac{1}{2}ln x + ln(x+1)$

$\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}$

$\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}$

$\frac{dy}{dx}= y(\frac{1}{2x} + \frac{1}{x+1})$

$\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{1}{2x} + \frac{1}{x+1})$

$\frac{dy}{dx}= \frac{\sqrt{x(x+1)}}{2x} + \frac{\sqrt{x(x+1)}}{x+1}$

answer in book $\frac{2x+1}{2\sqrt{x(x+1))}}$

So, although you have found the mistake that lead to the wrong answer I still have to ask: What is going on on lines 4 and 5?