Understanding the derivative of natural log function

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SUMMARY

The derivative of the natural logarithm function, ln(x), is definitively 1/x, while the derivative of a constant like ln(4) is 0. This distinction arises because ln(4) is a constant value, leading to a gradient of zero. The confusion often stems from misinterpreting constants as variables; however, the algebraic expression F'(x) = ln(x) results from applying the chain rule, where the derivative of a function is evaluated at a specific point rather than across its entire graph.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly derivatives
  • Familiarity with the natural logarithm function and its properties
  • Knowledge of the chain rule in differentiation
  • Ability to interpret graphical representations of functions
NEXT STEPS
  • Study the application of the chain rule in differentiation
  • Learn about the properties of logarithmic functions, including their derivatives
  • Explore graphical interpretations of derivatives and their significance
  • Investigate the concept of constants versus variables in calculus
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Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of logarithmic differentiation and its applications.

m3dicat3d
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Hope this is in the right place... I'm trying to understand why the derivative of ln(x) is 1/x while the derivative of something like ln(4) is 0. My knee-jerk reaction is to view 4 as representative of x, thereby giving me F'(x) ln(4) = 1/4, not 0. That would be the case, except ln(4) is a constant. Since I understand that ln(4) is a constant, the derivative should in fact be a zero. So maybe what is confusing me is why do we have the algebraic version of F'(x) ln(x) = 1/x in the first place (unless x represents another function instead of a number)?
 
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Re: understanding the derivative of natural log function

m3dicat3d said:
Hope this is in the right place... I'm trying to understand why the derivative of ln(x) is 1/x while the derivative of something like ln(4) is 0. My knee-jerk reaction is to view 4 as representative of x, thereby giving me F'(x) ln(4) = 1/4, not 0. That would be the case, except ln(4) is a constant. Since I understand that ln(4) is a constant, the derivative should in fact be a zero. So maybe what is confusing me is why do we have the algebraic version of F'(x) ln(x) = 1/x in the first place (unless x represents another function instead of a number)?

Imagine the derivative as being the gradient of an expression as per it's first principles. The graph of a the constant $y=a$ is a horizontal line so it has a gradient of $\dfrac{0}{\Delta x} = 0$ so long as $\Delta x \neq 0$.

When you take the derivative of $f(x)$ w.r.t $x$ at a given point P you're evaluating the gradient of f(x) at P, it's a point on a graph rather than a whole graph itself.

Strictly speaking of course the derivative of the natural logarithm is $\dfrac{f'(x)}{f(x)}$ because of the chain rule.
 
Re: understanding the derivative of natural log function

Thanks for the reply, I just starting writing out a long winded reply as to what I didn't understand, and in the process I thought through the matter more and ended up understanding it haha!

[moderator edit] The discussion regarding the posting issue has been moved here:

http://www.mathhelpboards.com/f25/unable-enter-carriage-return-into-my-posts-4787/
 
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