# Understanding the derivative of natural log function

#### m3dicat3d

##### New member
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
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
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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