# Using logarithmic differentiation

1. Oct 28, 2007

### rteng

using logarithmic differentiation I can get it to a point where I have...

ln(y)=20[ln(x+sqrt(x))-ln(x^2-2^x)]

I do not know what to do at that point

2. Oct 28, 2007

### rocomath

just take the derivative as you normally would

also you will have $$\frac{y'}{y}= ...$$

bring Y to the other side and plug in what Y is. the rest is straight forward differentiation.

3. Oct 28, 2007

### rteng

yeah but then you have a term that is:

[1/(x^2-2^x)]*(2x-d/dy2^x)

thats where I get stuck

4. Oct 28, 2007

### rocomath

i'm not really following

you mean on the $$x^{2}-2^{x}$$ part? how would you take it's derivative?

5. Oct 28, 2007

### rteng

$$y=((x+sqrt(x))/x^{2}-2^{x})^{20}$$

$$lny=20(ln(x+sqrt(x))-ln(x^{2}-2^{x}))$$

$$y'/y=(20/(x+sqrt(x))*(1+1/2(x)^{-1/2}))-(20/x^{2}-2^{x})*(2x-derivative of 2^x)$$

6. Oct 28, 2007

### rteng

yes that part sorry