- #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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SUMMARY
This discussion focuses on using logarithmic differentiation to simplify the differentiation of the function \( y = \left(\frac{x + \sqrt{x}}{x^2 - 2^x}\right)^{20} \). The key steps involve taking the natural logarithm of both sides, leading to the equation \( \ln(y) = 20[\ln(x + \sqrt{x}) - \ln(x^2 - 2^x)] \). Participants emphasize the importance of applying the product and quotient rules correctly, particularly when differentiating the term \( x^2 - 2^x \), which requires the derivative of \( 2^x \) to be included in the calculations.
PREREQUISITES- Understanding of logarithmic differentiation
- Familiarity with derivatives of exponential functions
- Knowledge of product and quotient rules in calculus
- Basic algebraic manipulation skills
- Practice logarithmic differentiation with various functions
- Learn how to differentiate exponential functions, specifically \( 2^x \)
- Review the product and quotient rules in calculus
- Explore applications of logarithmic differentiation in solving complex derivatives
Students and educators in calculus, mathematicians dealing with complex functions, and anyone looking to enhance their differentiation techniques using logarithmic methods.
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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