jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
Hi,
I am trying to work through a problem and think I have got part way through it but am not really sure what to do after a certain point. Could anyone give me a pointer?
\(\displaystyle y = ln(e^x(\frac{x-2}{x+2})^{\frac{3}{4}})\)
\(\displaystyle y =ln(u)\) , where u = \(\displaystyle e^x(\frac{x-2}{x+2})^{\frac{3}{4}}\)
\(\displaystyle \frac{dy}{du} = \frac{1}{u} = e^{-x}(\frac{x-2}{x+2})^{\frac{-3}{4}}\)
\(\displaystyle \frac{du}{dx} = e^x \left[\frac{3}{4}(\frac{x-2}{x+2})^\frac{-1}{4} \cdot (\frac{(x+2)-(x-2)}{(x+2)^2})\right] + e^x (\frac{x-2}{x+2})^{\frac{3}{4}}\)
\(\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
So that gives:
\(\displaystyle e^{-x}(\frac{x-2}{x+2})^{\frac{-3}{4}}\cdot e^x \left[\frac{3}{4}(\frac{x-2}{x+2})^{\frac{-1}{4}}\cdot(\frac{(x+2)-(x-2)}{(x+2)^2})\right]+e^x(\frac{x-2}{x+2})^{\frac{3}{4}}\)
\(\displaystyle = e^{-x}(\frac{x-2}{x+2})^{\frac{-3}{4}}\cdot e^x \left[\frac{3}{4}(\frac{x-2}{x+2})^{\frac{-1}{4}}\cdot(\frac{4}{(x+2)^2})\right]+e^x(\frac{x-2}{x+2})^{\frac{3}{4}}\)
I am not sure how to proceed with breaking this down though. I've checked and the answer I am aiming for is \(\displaystyle \frac {x^2-1}{x^2-4}\)
I am trying to work through a problem and think I have got part way through it but am not really sure what to do after a certain point. Could anyone give me a pointer?
\(\displaystyle y = ln(e^x(\frac{x-2}{x+2})^{\frac{3}{4}})\)
\(\displaystyle y =ln(u)\) , where u = \(\displaystyle e^x(\frac{x-2}{x+2})^{\frac{3}{4}}\)
\(\displaystyle \frac{dy}{du} = \frac{1}{u} = e^{-x}(\frac{x-2}{x+2})^{\frac{-3}{4}}\)
\(\displaystyle \frac{du}{dx} = e^x \left[\frac{3}{4}(\frac{x-2}{x+2})^\frac{-1}{4} \cdot (\frac{(x+2)-(x-2)}{(x+2)^2})\right] + e^x (\frac{x-2}{x+2})^{\frac{3}{4}}\)
\(\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
So that gives:
\(\displaystyle e^{-x}(\frac{x-2}{x+2})^{\frac{-3}{4}}\cdot e^x \left[\frac{3}{4}(\frac{x-2}{x+2})^{\frac{-1}{4}}\cdot(\frac{(x+2)-(x-2)}{(x+2)^2})\right]+e^x(\frac{x-2}{x+2})^{\frac{3}{4}}\)
\(\displaystyle = e^{-x}(\frac{x-2}{x+2})^{\frac{-3}{4}}\cdot e^x \left[\frac{3}{4}(\frac{x-2}{x+2})^{\frac{-1}{4}}\cdot(\frac{4}{(x+2)^2})\right]+e^x(\frac{x-2}{x+2})^{\frac{3}{4}}\)
I am not sure how to proceed with breaking this down though. I've checked and the answer I am aiming for is \(\displaystyle \frac {x^2-1}{x^2-4}\)
