# logarithmic differentiation

• May 15th 2010, 12:49 PM
cdlegendary
logarithmic differentiation
how do you take the derivative of $y=((x^2+7)/(x^2+8))^{1/7}$?

I used logarithmic differentiation and got: $(((2x)/(7x^2+49))-((2x)/(7x^2+56)))((x^2+7)/(x^2+8))^{1/7}$

Doesn't seem to be correct, though..
• May 15th 2010, 01:27 PM
apcalculus
Have you tried checking your work in WolframAlpha?

http://www.wolframalpha.com/input/?i=differentiate+y%3D((x^2%2B7)/(x^2%2B8))^{1/7

Good luck!
• May 15th 2010, 03:10 PM
HallsofIvy
Quote:

Originally Posted by cdlegendary
how do you take the derivative of $y=((x^2+7)/(x^2+8))^{1/7}$?

I used logarithmic differentiation and got: $(((2x)/(7x^2+49))-((2x)/(7x^2+56)))((x^2+7)/(x^2+8))^{1/7}$

Doesn't seem to be correct, though..

$ln(y)= ln(x^2+ 7)- (1/7)ln(x^2+ 8)$
so $\frac{1}{y}\frac{dy}{dx}= \frac{2x}{x^2+7}- \frac{2x}{7(x^2+ 8)}$.

I cannot see any reason to have the "7" multiplied in the denominator of the first fraction.
• May 15th 2010, 10:10 PM
cdlegendary
derivative
http://hw.math.ucsb.edu/webwork/math...834855img1.gif

i need to take the derivative of the above equation...none of the results seem to be working. and wolfram alpha gets a strange result, too.
• May 15th 2010, 10:29 PM
Prove It
Quote:

Originally Posted by HallsofIvy
$ln(y)= ln(x^2+ 7)- (1/7)ln(x^2+ 8)$
so $\frac{1}{y}\frac{dy}{dx}= \frac{2x}{x^2+7}- \frac{2x}{7(x^2+ 8)}$.

I cannot see any reason to have the "7" multiplied in the denominator of the first fraction.

Actually since

$y = \left(\frac{x^2 + 7}{x^2 + 8}\right)^{\frac{1}{7}}$

That means

$\ln{y} = \ln{\left(\frac{x^2 + 7}{x^2 + 8}\right)^{\frac{1}{7}}}$

$= \frac{1}{7}\ln{\left(\frac{x^2 + 7}{x^2 + 8}\right)}$

$= \frac{1}{7}[\ln{(x^2 + 7)} - \ln{(x^2 + 8)}]$

$= \frac{1}{7}\ln{(x^2 + 7)} - \frac{1}{7}\ln{(x^2 + 8)}$.

Therefore

$\frac{1}{y}\,\frac{dy}{dx} = \frac{2x}{7(x^2 + 7)} - \frac{2x}{7(x^2 + 8)}$

$= \frac{2x(x^2 + 8) - 2x(x^2 + 7)}{7(x^2 + 7)(x^2 + 8)}$

$= \frac{2x(x^2 + 8 - x^2 - 7)}{7(x^2 + 7)(x^2 + 8)}$

$= \frac{2x}{7(x^2 + 7)(x^2 + 8)}$.

Therefore

$\frac{dy}{dx} = \frac{2xy}{7(x^2 + 7)(x^2 + 8)}$

$= \frac{2x\left(\frac{x^2 + 7}{x^2 + 8}\right)^{\frac{1}{7}}}{7(x^2 + 7)(x^2 + 8)}$

$= \frac{\frac{2x(x^2+7)^{\frac{1}{7}}}{(x^2 + 8)^{\frac{1}{7}}}}{7(x^2 + 7)(x^2 + 8)}$

$= \frac{2x(x^2 + 7)^{\frac{1}{7}}}{7(x^2 + 7)(x^2 + 8)(x^2 + 8)^{\frac{1}{7}}}$

$= \frac{2x}{7(x^2 + 7)^{\frac{6}{7}}(x^2+ 8)^{\frac{8}{7}}}$.