How does the chain rule apply to logarithmic differentiation?

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SUMMARY

The discussion focuses on the application of the chain rule in logarithmic differentiation, specifically how to differentiate the natural logarithm of a function y(x). The key equation derived is d/dx(ln y(x)) = (1/y(x)) * (dy/dx), which simplifies to y' = (y * d/dx(ln y)). Participants clarify that the chain rule requires differentiating the outer function (ln y) first, followed by the inner function (y(x)). This understanding is essential for correctly applying logarithmic differentiation in calculus.

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  • Understanding of basic calculus concepts, particularly derivatives.
  • Familiarity with the chain rule in differentiation.
  • Knowledge of logarithmic functions and their properties.
  • Ability to manipulate algebraic expressions involving functions.
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  • Study the chain rule in-depth, focusing on its application in various differentiation scenarios.
  • Practice logarithmic differentiation with different functions to solidify understanding.
  • Explore advanced calculus topics such as implicit differentiation and its relationship with logarithmic differentiation.
  • Review examples of real-world applications of logarithmic differentiation in fields like physics and engineering.
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Students in introductory calculus courses, educators teaching differentiation techniques, and anyone seeking to strengthen their understanding of logarithmic differentiation and the chain rule.

Owen-
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Hi everyone - I'm in my first year at uni, and I was given a derivative that i don't quite get:

y(x) is a function of x then
y` = (dy/dx)

d/dx (ln y(x))=1/y(x)*(dy/dx) =(y`/y)


This is given by the chain rule. I don't understand this step...

Any help?

Thanks in advance,
Owen.
 
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chain rule says deriver the outside function first (the ln y part)
then do the derivative of the inside function (the y(x))
d/dx(ln y(x))
= d/dx(ln y)*d/dx (y(x))
=1/y *y'(x)
 
Thanks, just needed to have that put another way - makes perfect sense - thanks again
 

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