Differentiating a^x from first principles

[kdog72]

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 a^x is ln(a)*a^x 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 a^x from first principles, but I got stuck.

From lim h->0 ((a^x+h - a^x)/h) i got: a^x lim h->0 ((a^h - 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 a^h, 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 ((a^h - 1)/h)

Thanks

 

[Jason76]

I know that the derivative of a^x is ln(a)*a^x but I wanted to try work it out from first principles

The derivative of $\displaystyle a^{x}$ is $\displaystyle a^{x} * dx * \ln a$

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

 

[eddybob123]

$\displaystyle y=a^x$

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

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

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

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

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

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

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

 

[kdog72]

$\displaystyle y=a^x$

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

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

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

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

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

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

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

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 a^x)

Thanks anyway

 

[pka]

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 a^x)

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

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

 

[HallsofIvy]

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 a^x)

Thanks anyway

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

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 $\displaystyle f'(x)= C_a a^x$, a constant times the original function. How we determine what "$\displaystyle C_a$" 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, $\displaystyle C_1= 0$, if a= 2, the limit exists and $\displaystyle C_2$ is less than 1 while if a= 3, $\displaystyle C_3$ exists and is larger than 1. That means that there exist a value of a, between 2 and 3, such that $\displaystyle C_a= 1$! Narrowing it further, we can see that $\displaystyle C_{2.7}< 1$ while $\displaystyle C_{2.8}> 1$, so that value must be between 2.7 and 2.8, that $\displaystyle C_{2.72}> 1$ 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 $\displaystyle C_a= 1$. We call that number "e" so that $\displaystyle C_e= 1$ and $\displaystyle (e^x)'= (1)(e^x)= e^x$. We can then use the fact that $\displaystyle a^x= e^{ln(a^x)}= e^{aln(x)}$ and the chain rule to show that, for any a>1, $\displaystyle C_a= ln(x)$ and $\displaystyle (a^x)'= ln(a) a^x$.

Editted thanks to look again.

 

[lookagain]

... so that value must be between 2.7 and 2.8, that $\displaystyle C_{2.72}> 1$ 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 $\displaystyle C_a= 1$

Typo. $\displaystyle \ \$ It should be "2.7..."



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


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



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



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





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



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





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




$\displaystyle ln(a)$