# Thread: Differentiating a^x from first principles

1. ## Differentiating a^x from first principles

Hi

I'm new to this place so I guess I should say I'm from NZ and in first year of university. I'm not actually doing maths this year, but I did university-level maths during my last year of high school (so just assume I have the knowledge of a university first-year student), and maths is still something of intense interest.

So, to the problem: I know that the derivative of ax is ln(a)*ax but I wanted to try work it out from first principles

I've tried searching the internet for answers, but nothing has come up. So I was trying to differentiate ax from first principles, but I got stuck.

From lim h->0 ((ax+h - ax)/h) i got: ax lim h->0 ((ah - 1)/h) but I couldn't get any further. You end up with a '0/0' situation which if I remember correctly, you can use L'Hopital's rule, but since it has ah, the derivative of which is what I'm originally trying to work out, that doesn't seem to work.

So, my question is: how do I work out lim h->0 ((ah - 1)/h)

Thanks

2. I know that the derivative of ax is ln(a)*ax but I wanted to try work it out from first principles
The derivative of $a^{x}$ is $a^{x} * dx * \ln a$

or derivative of $a^{u}$ is $a^{u} * du * \ln a$

3. $y=a^x$

$\ln(y)=\ln(a^x)$

$\ln(y)=x\ln(a)$

$\frac{d}{dx}\ln(y)=\frac{d}{dx}x\ln(a)$

$\frac{d}{dy}\ln(y)\times\frac{dy}{dx}=\ln(a)$

$\frac{1}{y}\times\frac{dy}{dx}=\ln(a)$

$\frac{dy}{dx}=y\ln(a)$

$\frac{dy}{dx}=a^x\ln(a)$

4. Originally Posted by eddybob123
$y=a^x$

$\ln(y)=\ln(a^x)$

$\ln(y)=x\ln(a)$

$\frac{d}{dx}\ln(y)=\frac{d}{dx}x\ln(a)$

$\frac{d}{dy}\ln(y)\times\frac{dy}{dx}=\ln(a)$

$\frac{1}{y}\times\frac{dy}{dx}=\ln(a)$

$\frac{dy}{dx}=y\ln(a)$

$\frac{dy}{dx}=a^x\ln(a)$
I understand that, but what I'm looking for is a solution using first principles (i.e. from lim h->0 ((f(x+h) - f(x))/h) where in this case f(x) is ax)

Thanks anyway

5. Originally Posted by kdog72
I understand that, but what I'm looking for is a solution using first principles (i.e. from lim h->0 ((f(x+h) - f(x))/h) where in this case f(x) is ax)
I do understand and appreciate your question. But I fear you will be disappointed with the actual answer.
At some point it simply comes down to a matter of definitions.

For example, Leonard Gillman defines $x > 0,\;\log (x) =\displaystyle \int_1^x {\frac{1}{t}dt}$.

From that it is possible( not easy) to show that $\displaystyle{\lim _{h \to 0}}\frac{{{a^h} - 1}}{h} = \log (a)$.

6. Originally Posted by kdog72
I understand that, but what I'm looking for is a solution using first principles (i.e. from lim h->0 ((f(x+h) - f(x))/h) where in this case f(x) is ax)

Thanks anyway
From "first principles" (that is, from the definition, $f'(x)= \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$) would be:
$f(x+h)- f(x)= a^{x+h}- a^x= a^xa^h- a^x= a^x(a^h- 1)$
so $\frac{f(x+h)- f(x)}{h}= \frac{a^x(a^h- 1)}{h}= \left[\frac{a^h- 1}{h}\right]a^x$
and then $f'(x)= \left[\lim_{h\to 0}\frac{a^h- 1}{h}\right]a^x$.

At this point we can see that the quantity inside the braces depends on a but NOT the variable, x, while the quantity outside has no "h". That means that the limit, as h goes to 0 (assuming it exists) is a constant depending on a but not the variable x, so we can write $f'(x)= C_a a^x$, a constant times the original function. How we determine what "$C_a$" is, depends on how we define things. It is, of course, ln(a) but how, exactly, do we define that?

One method is this: by experimenting with different values, it is easy to see that if a= 1, $C_1= 0$, if a= 2, the limit exists and $C_2$ is less than 1 while if a= 3, $C_3$ exists and is larger than 1. That means that there exist a value of a, between 2 and 3, such that $C_a= 1$! Narrowing it further, we can see that $C_{2.7}< 1$ while $C_{2.8}> 1$, so that value must be between 2.7 and 2.8, that $C_{2.72}> 1$ so that value must be between 2.70 and 2.72, etc. The result of all this is that there exist a number, a, around 2.7... such that $C_a= 1$. We call that number "e" so that $C_e= 1$ and $(e^x)'= (1)(e^x)= e^x$. We can then use the fact that $a^x= e^{ln(a^x)}= e^{aln(x)}$ and the chain rule to show that, for any a>1, $C_a= ln(x)$ and $(a^x)'= ln(a) a^x$.

Editted thanks to look again.

7. Originally Posted by HallsofIvy
... so that value must be between 2.7 and 2.8, that $C_{2.72}> 1$ so that value must be between 2.70 and 2.72, etc.

The result of all this is that there exist a number, a, around > > > 2.1... < < < such that $C_a= 1$
Typo. $\ \$ It should be "2.7..."

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Suppose we know that $\ e^x \ = \ 1 \ + \ x \ + \ \dfrac{x^2}{2!} \ + \ \dfrac{x^3}{3!} \ + \ \dfrac{x^4}{4!} \ + \ ...$

and that $\ a^h \ = \ e^{h*ln(a)}, \ \$ where $\$ ln(a) = $\ \log_e(a).$

For this portion of the problem, show that $\ \displaystyle\lim_{h\to 0} \ \dfrac{a^h \ - \ 1}{h} \ = \ ln(a). \ \ \ \ \ \ \ \$ (There are appropriate restrictions on a.)

$\displaystyle\lim_{h\to 0} \ \dfrac{\bigg( 1 \ + \ h*ln(a) \ + \ \dfrac{[h*ln(a)]^2}{2!} \ + \ \dfrac{[h*ln(a)]^3}{3!} \ + \ \dfrac{[h*ln(a)]^4}{4!} \ + \ ... \ \bigg) \ - \ 1 \ }{h} \ =$

$\displaystyle\lim_{h\to 0} \ \dfrac{ \ h*ln(a) \ + \ \dfrac{[h*ln(a)]^2}{2!} \ + \ \dfrac{[h*ln(a)]^3}{3!} \ + \ \dfrac{[h*ln(a)]^4}{4!} \ + \ ... \ \ }{h} \ =$

$\displaystyle\lim_{h\to 0} \ \dfrac{ \ h*ln(a) \ + \ \dfrac{h^2*[ln(a)]^2}{2!} \ + \ \dfrac{h^3*[ln(a)]^3}{3!} \ + \ \dfrac{h^4*4ln(a)]^4}{4!} \ + \ ... \ \ }{h} \ =$

$\displaystyle\lim_{h\to 0} \ \bigg[ ln(a) \ + \ \dfrac{h*[ln(a)]^2}{2!} \ + \ \dfrac{h^2*[ln(a)]^3}{3!} \ + \ \dfrac{h^3*[ln(a)]^4}{4!} \ + \ ... \bigg] \ =$

$ln(a)$