logarithmic differentiation

Jan 2010
79
1
how do you take the derivative of \(\displaystyle y=((x^2+7)/(x^2+8))^{1/7} \)?

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

Doesn't seem to be correct, though..
 
Apr 2009
293
94
Boston
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!
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
how do you take the derivative of \(\displaystyle y=((x^2+7)/(x^2+8))^{1/7} \)?

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

Doesn't seem to be correct, though..
\(\displaystyle ln(y)= ln(x^2+ 7)- (1/7)ln(x^2+ 8)\)
so \(\displaystyle \frac{1}{y}\frac{dy}{dx}= \frac{2x}{x^2+7}- \frac{2x}{7(x^2+ 8)}\).

I cannot see any reason to have the "7" multiplied in the denominator of the first fraction.
 
Jan 2010
79
1
derivative



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.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
\(\displaystyle ln(y)= ln(x^2+ 7)- (1/7)ln(x^2+ 8)\)
so \(\displaystyle \frac{1}{y}\frac{dy}{dx}= \frac{2x}{x^2+7}- \frac{2x}{7(x^2+ 8)}\).

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

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

That means

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

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

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

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


Therefore

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

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

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

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


Therefore

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

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

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

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

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