1. ## Derivative at f1(0)

If you use the limit method for calculating a derivative, the denominator is the difference between x and (a) as (a) approaches x. But that limit would be equal to zero. So how is it that you can actually find the derivative since you would have to divide by zero?

Did that make sense?

2. Originally Posted by CrazyAsian
If you use the limit method for calculating a derivative, the denominator is the difference between x and (a) as (a) approaches x. But that limit would be equal to zero. So how is it that you can actually find the derivative since you would have to divide by zero?

Did that make sense?
if a function is differentiable, there will be a way to factorize the top, so you cancel something in the bottom that makes the bottom zero. try it with a function you know, say x^2

and it's the limit as x approaches a

3. The function I was thinking about was a^x. Sorry, I'm sure there is a way to factorize it but my math is very very rusty.

4. Did that make sense?
No

5. Originally Posted by CrazyAsian
The function I was thinking about was a^x. Sorry, I'm sure there is a way to factorize it but my math is very very rusty.
You're trying to use the definition of the derivative to find the derivative of $\displaystyle f(x)=a^x$?

6. Originally Posted by ecMathGeek
You're trying to use the definition of the derivative to find the derivative of $\displaystyle f(x)=a^x$?
yea, i made two attempts and gave up

i tried using both (equivalent) definitions:

$\displaystyle f'(b) = \lim_{x \to b} \frac {f(x) - f(b)}{x - b}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$

to no avail

7. Originally Posted by ecMathGeek
You're trying to use the definition of the derivative to find the derivative of $\displaystyle f(x)=a^x$?
Yes. Am I missing something really obvious?

8. Originally Posted by Jhevon
yea, i made two attempts and gave up

i tried using both (equivalent) definitions:

$\displaystyle f'(b) = \lim_{x \to b} \frac {f(x) - f(b)}{x - b}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$

to no avail
I did too. Sadly, not matter what method I use, I tend to go back to wanting to use L'Hopital's rule (which requires me to already know the derivative of $\displaystyle a^x$). lol.

I don't know if this derivative was ever actually shown to me using the definition of the derivative.

CrazyAsian, I don't think you're missing anything obvious becuase if so, Jhevon and I are missing that same thing.

9. Btw, how do you guys write nicely formatted equations like that?

10. Originally Posted by CrazyAsian
Btw, how do you guys write nicely formatted equations like that?
we use LaTex, its easy to learn, see the tutorial here

11. If $\displaystyle a>1$ then $\displaystyle \lim _{h \to 0} \frac{{a^{x + h} - a^x }}{h} = a^x \lim _{h \to 0} \frac{{a^h - 1}}{h}$.

Now the $\displaystyle \lim _{h \to 0} \frac{{a^h - 1}}{h}$ is called by Serge Lang the mysterious limit. There is a good but informal discussion of that fact in Lang’s A First Course in Calculus. This is by no means easy to prove. In beginning courses, it is discussed informally.

12. i think i got something. it may be a bit choppy though.

Recall that $\displaystyle \frac {d}{dx} e^u = \frac {d}{dx} u \cdot e^u$ where $\displaystyle u$ is a function of $\displaystyle x$

Note that: $\displaystyle a^x = e^{\ln a^x} = e^{x \ln a}$

Now consider $\displaystyle f(x) = a^x$

$\displaystyle \Rightarrow \frac {d}{dx} a^x = \frac {d}{dx} e^{x \ln a}$

$\displaystyle = \frac {d}{dx} x \ln a \cdot e^{x \ln a}$

$\displaystyle = \ln a \cdot \frac {d}{dx} x \cdot e^{x \ln a}$

Now apply the definition:

$\displaystyle \Rightarrow \frac {d}{dx} a^x = \ln a \cdot \lim_{h \to 0} \frac {(x + h) - x}{h} \cdot e^{x \ln a}$

$\displaystyle \Rightarrow \frac {d}{dx} a^x = \ln a \cdot e^{x \ln a} = \ln a \cdot a^x$

The reason i do not like this though is that i think the relationship $\displaystyle \frac {d}{dx} e^u = \frac {d}{dx} u \cdot e^u$ only works if $\displaystyle u$ is a linear function of $\displaystyle x$

EDIT: Nevermind, $\displaystyle x \ln a$ is a linear function! since $\displaystyle a$ is a constant. So what do you guys think?

13. As long as $\displaystyle a$ is a constant, $\displaystyle \frac {d}{dx} e^u = \frac {d}{dx}$\displaystyle u$\cdot e^u$ is true. u does not have to be a linear function.

The problem I had with doing this is that we would have to show using the definition of the derivative $\displaystyle \frac {d}{dx} e^u = \frac {d}{dx} u \cdot e^u$ is true.

14. Originally Posted by ecMathGeek
As long as $\displaystyle a$ is a constant, $\displaystyle \frac {d}{dx} e^u = \frac {d}{dx} u \cdot e^u$ is true. u does not have to be a linear function.
good to know

The problem I had with doing this is that we would have to show using the definition of the derivative $\displaystyle \frac {d}{dx} e^u = \frac {d}{dx} u \cdot e^u$ is true.
yea, but we have to draw the line somewhere

i just thought that $\displaystyle \frac {d}{dx} e^u = \frac {d}{dx} u \cdot e^u$ would be a part of the background knowledge of someone wanting to prove $\displaystyle \frac {d}{dx} a^x = \ln a \cdot a^x$...even though i didn't know that fact as well as i should. so i kind of treated that fact as a proven lemma that didn't need to be proven again. i guess we could prove it before though...do i know how to prove that?

15. Using the definition is the bad way to do it.
For example,
$\displaystyle \lim n^{1/n} = 1$
Try doing that with the defintion!!

Instead, we use more powerful techiniques which are proven by the definitions.

$\displaystyle f(x)=a^x \mbox{ for }a> 0$ is by definition $\displaystyle f(x) = e^{x\ln a}$. We can find the derivative of this using the chain rule since $\displaystyle x\ln a$ is differenciable as $\displaystyle e^x$ is differenciable everywhere. So $\displaystyle f'(x) = \ln a e^{\ln a x} = \ln a a^x$.

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