# What is dy/dx of a square root using logarithmic differentiation?

I thought I knew how to do logarithmic differentiation until I came across this:

$y = \sqrt{x(x+7)}$ find dy/dx by logarithmic differentiation

I was not phased by the problem and did the following work:

$\ln(y) = \ln(\sqrt{x(x+7)})$

$\ln(y) = {1\over 2}(\ln(x) + \ln(x+7))$

${1\over y}{dy\over dx} = {1\over 2}({1\over x}+{1\over x+7})$

${dy\over dx} = {y\over 2}({2x+7\over x^2+7x})$

but then found out that that was not actually what the answer was if you take dy/dx without logarithmic differentiation. What am I doing wrong? I have spent the past hour trying to figure it out with no luck.

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Your result is correct: tinyurl.com/9s8dwsv – Eric Angle Oct 18 '12 at 5:07

Looks to me like you've done it correctly. Since $y = \sqrt{x^2 + 7x}$, $$\frac{y}{2}\left(\frac{2x + 7}{x^2 + 7x}\right) = \frac{1}{2}\frac{(2x + 7)\sqrt{x^2 + 7x}}{x^2 + 7x} = \frac{2x + 7}{2\sqrt{x^2 + 7x}}.$$ This last is of course the derivative (after chain rule) of $\sqrt{x^2 + 7x}$.
You did everything perfectly. Resubstitute $$y = \sqrt{x(x+7)}$$ to get $$y' = \frac{\sqrt{x(x+7)}}{2}\cdot\frac{2x+7}{x(x+7)} = \frac{2x+7}{2\sqrt{x(x+7)}}$$ This is precisely what I get by applying normal differentiation rules.