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Why do we use the natural log in the derivative of an exponential function?

by QuickLoris
Tags: derivative, exponential, function, natural
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QuickLoris
#1
Jan22-13, 02:38 PM
P: 12
I recently struck a question that I have not been able to find an answer to. I feel like I'm missing something obvious, so I've come here for help.

The derivative of [itex]a^{x}[/itex] is [itex]a^{x}[/itex]lna.

The explanation that Stewart 5e gives is:
[itex]\frac{d}{dx}[/itex][itex]a^{x}[/itex] = [itex]\frac{d}{dx}[/itex][itex]e^{(lna)x}[/itex]

= [itex]e^{(lna)x}[/itex][itex]\frac{d}{dx}[/itex](lna)x

=[itex]e^{(lna)x}[/itex][itex]\cdot[/itex]lna

=[itex]a^{x}[/itex]lna

My question is: Why do we use the natural log instead of a log of any other base?
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mathman
#2
Jan22-13, 03:30 PM
Sci Advisor
P: 6,040
ax = e(lna)x

d/dx(ecx) = cecx

If you used any base other than e, the second equation would be a problem.
QuickLoris
#3
Jan22-13, 03:58 PM
P: 12
So, in other words, since e is defined so that lim [itex]e^{h}[/itex]=1 as h[itex]\rightarrow[/itex]0, the derivative is itself. Otherwise, the derivative would be recursive? as in,

f(x) = [itex]a^{x}[/itex]

[itex]\frac{d}{dx}[/itex]f(x) = [itex]a^{x}[/itex][itex]\frac{d}{dx}[/itex]f(0)

Is that right?

chiro
#4
Jan22-13, 09:23 PM
P: 4,572
Why do we use the natural log in the derivative of an exponential function?

Hey QuickLoris and welcome to the forums.

The natural base has so many properties for so many applications including pure mathematics, applied mathematics, and statistics, that it is just well suited for these things and as such it becomes not only a tool of frequent use, but also one of investigation.

You have for example the connection between the trig functions to the hyperbolic ones and the exponential via Eulers formula and the complex valued analogs for the trig and hyperbolic.

In statistics you have probability transform functions, distributions, and a variety of other things involving the exponential function.

There are just so many connections that it becomes kind of a "neat coincidence" for all of mathematics.
lurflurf
#5
Jan23-13, 12:07 AM
HW Helper
P: 2,263
Use any base you like

$$\dfrac{d}{dx}a^x=\frac{\log_b(a)}{\log_b(e)} a^x$$

We can see if b=a or e, we will only need one log.
HallsofIvy
#6
Jan23-13, 06:56 AM
Math
Emeritus
Sci Advisor
Thanks
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P: 39,363
Quote Quote by QuickLoris View Post
So, in other words, since e is defined so that lim [itex]e^{h}[/itex]=1 as h[itex]\rightarrow[/itex]0, the derivative is itself. Otherwise, the derivative would be recursive? as in,

f(x) = [itex]a^{x}[/itex]

[itex]\frac{d}{dx}[/itex]f(x) = [itex]a^{x}[/itex][itex]\frac{d}{dx}[/itex]f(0)

Is that right?
No, it is not right. Since f(0) is a number, a constant, and does not depend on x, "df(0)/dx" is equal to 0.
QuickLoris
#7
Jan23-13, 12:50 PM
P: 12
Quote Quote by lurflurf View Post
Use any base you like

$$\dfrac{d}{dx}a^x=\frac{\log_b(a)}{\log_b(e)} a^x$$

We can see if b=a or e, we will only need one log.
I understand that you can use the change of base formula to to change the base to whatever you like once you have the derivative, I just wanted to know why ln was chosen to begin with. mathman somewhat answered my question.

Quote Quote by HallsofIvy View Post
No, it is not right. Since f(0) is a number, a constant, and does not depend on x, "df(0)/dx" is equal to 0.
I should have used different notation. I mean f '(0), not f(0).


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