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Understanding the derivative of natural log function

m3dicat3d

New member
Jan 11, 2013
19
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)?
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
Re: understanding the derivative of natural log function

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.
 

m3dicat3d

New member
Jan 11, 2013
19
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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