- #1
rteng
- 26
- 0
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
Using logarithmic differentiation
- Thread starter rteng
- Start date
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Homework Help Overview
The discussion revolves around the application of logarithmic differentiation in the context of a function involving both polynomial and exponential components. Participants are exploring the differentiation process after applying logarithmic properties to simplify the expression.
Discussion Character
- Exploratory, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- Participants discuss the steps following the logarithmic differentiation, including taking derivatives and manipulating terms. Questions arise regarding the differentiation of specific components, particularly the term involving \(x^2 - 2^x\) and its derivative.
Discussion Status
The discussion is ongoing, with participants providing insights into the differentiation process. Some guidance has been offered regarding the general approach to taking derivatives, but there remains uncertainty about specific terms and how to proceed from the current point in the differentiation.
Contextual Notes
Participants are grappling with the complexity of the derivative involving both polynomial and exponential functions, and there is a noted lack of clarity on how to handle the derivative of \(2^x\) in the context of the overall expression.
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
rocomath
- 1,752
- 1
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.
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
rteng
- 26
- 0
yeah but then you have a term that is:
[1/(x^2-2^x)]*(2x-d/dy2^x)
thats where I get stuck
[1/(x^2-2^x)]*(2x-d/dy2^x)
thats where I get stuck
- #4
rocomath
- 1,752
- 1
i'm not really following
you mean on the x^{2}-2^{x} part? how would you take it's derivative?
you mean on the x^{2}-2^{x} part? how would you take it's derivative?
- #5
rteng
- 26
- 0
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)
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
rteng
- 26
- 0
yes that part sorry
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