# logarithmic differentiation

#### cdlegendary

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..

#### 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!

#### HallsofIvy

MHF Helper
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.

#### cdlegendary

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
$$\displaystyle ln(y)= ln(x^2+ 7)- (1/7)ln(x^2+ 8)$$
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

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

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

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

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

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