Ways to find a derivative.

  • Limit of slopes of secant lines (in books).
  • First define the best linear approximation to the function (Frechet derivative), and then define the derivative as the slope of this line.
  • Use infinitesmals, by defining the hyper reals.
  • "The product of two small things is so small that we will ignore it", letting the derivative be the part that is not the product of two small things.

For the transcendental functions sine and cosine, we use the cosine addition formula if we go with the epsilon-delta approach. For the infinitesmal approach, we use a unit circle and note that cos(theta)=(sin(theta + dtheta)-sin(theta))/dtheta. From that picture, follows the derivatives (in a slightly nonrigorous way in Keisler's text). Perhaps using a non-rigorous approach is the way to go for non-transcendental functions.

Sometimes we need inductive logic, sometimes we need deductive logic. Which is appropriate? Maybe the key is to stop deciding which is appropriate, and instead start focusing on which we are using. If we learn to diagnose the type of logic used, we can identify what flaws may follow from it. If we are happy with the power rule, by doing a few examples, then we move on. Those who want to deductively finish the problem can do so on their own. Those who are happy with the inductive logic, can move on. Perhaps this is really the issue between making something "rigorous" or not. I do know that in my own problem sets, I sometimes leave out assumptions. We induce things all the time, but seldom do we attach appropriate conditions (nor do we address what they are). That is a flaw.

Equivalent Definintions

(Have to check domain issues on these)

  • For every $\varepsilon >0 $, there exists a $\delta>0$ such that if $0<|h|<\delta$ then $\ds \left|\frac{f(x+h)-f(x)}{h}-f'(x)\right| < \varepsilon$.
  • For every $\varepsilon >0 $, there exists a $\delta>0$ such that if $0<|\Delta x|<\delta$ then $\ds \left|f(x+\Delta x)-f(x)-f'(x)\Delta x\right| < |\Delta x| \varepsilon$.
  • There exists $q(\Delta x)$ and $r(\Delta x)$ so that $f(x+\Delta x)-f(x) = q(\Delta x)\cdot \Delta x + |\Delta x|r(\Delta x)$ such that $ \ds \lim_{\Delta x \to 0}q(\Delta x)$ exists and $ \ds \lim_{\Delta x \to 0}r(\Delta x) = 0$.

To get the last to work, I need to make sure that the equality holds on some small neighborhood (not just on their domains). I want to make sure that the equality actually holds. To do this, I'll also have to make sure that I appropriately define the derivative.... Then prove that each implies the other.

The fact that difference-quotient and fretchet match up is pretty standard.

  • Suppose (frechet) for every $\varepsilon >0 $, there exists a $\delta>0$ such that if $0<|\Delta x|<\delta$ then $\ds \left|f(x+\Delta x)-f(x)-f'(x)\Delta x\right| < |\Delta x| \varepsilon$. We must prove "We can write $f(x+\Delta x)-f(x) = m \cdot \Delta x + |\Delta x|\varepsilon $ where $ \ds \lim_{\Delta x \to 0}m$ exists and $ \ds \lim_{\Delta x \to 0}\varepsilon = 0$. " Let $q(\Delta x) = f'(x)$ and $r(\Delta x) = \frac{f(x+\Delta x)-f(x) - q(\Delta x)\cdot \Delta x}{|\Delta x|}$. Clearly

New option.

Let $f$ be a function (domain,range given). We will not define differentiability of $f$ at a point $x$ unless $x$ is an interior point of the domain. That said, provided $x$ is an interior point in the domain of $f$ (let $U$ be the required open set), we have three equivalent definitions:

  1. $\ds\lim_{h\to 0}\frac{f(x+h)-f(x) }{h}$ exists. In other words, There exists $L$ such that for every $\varepsilon>0$, there exists a $\delta >0$ such that for every $h$ with $x+h\in U$, we have if $0 < |h|<\delta$ then $|\frac{f(x+h)-f(x) }{h} - L| < \varepsilon$.
  2. There exists a linear transformation $m$ (so a function of the form $dy=m dx$) such that for every $\varepsilon>0$, there exists a $\delta >0$ such that for every $\Delta x$ with $x+\Delta x\in U$, we have if $0 < |\Delta x|<\delta$ then $|f(x+\Delta x)-f(x) - m\Delta x| < |\Delta x| \varepsilon$.
  3. For every $\Delta x\neq 0$ with $x+\Delta x\in U\setminus{x}$, there exists a linear transformation $q_x(\Delta x)$ and an expression $r_x(\Delta x)$ such that (1) q_x(\Delta x) has a limit (using operator norms) as $\Delta x\to 0$, (2) $r_x(\Delta x)\to 0$ as $\Delta x\to 0$ (codomain norm), and (3) we have $f(x+\Delta x)-f(x) = q_x(\Delta x)(\Delta x) +|\Delta x| r_x(\Delta x)$.
  4. There exists a linear transformation $q_x(\Delta x)$ and an expression $r_x(\Delta x)$ such that (1) $q_x(\Delta x)$ has a limit as $\Delta x\to 0$, (2) $r_x(\Delta x)/|\Delta x|\to 0$ as $\Delta x\to 0$, and (3) for every $\Delta x$ with $x+\Delta x\in U\setminus{x}$, we have $f(x+\Delta x)-f(x) = q_x(\Delta x)\Delta x +r_x(\Delta x)$.

Fun problem here. Note that for generalization, we need $q$ to be a linear transformation, and $r$ to be a vector in the codomain (they are different objects). So to deal with limits, we have to work with different types of limits (limits of linear transformations, and limits of vectors). This adds a huge cognitive load on the back end (if you really want to deal with the specifics), but I think it greatly simplifies things early on.

In the 1D case, both the function and the vector are single numbers, so the complexity is gone. In the 2D case, we can reduce both problems to regular 2D distances (as both work as norms on linear functions and vectors). For finite dimensional vector spaces, can we always use the L2 norm? My guess is YES. However, using absolute values might be easier in some cases...

I need to prove that the last line is equivalent to the other 2. It is common knowledge that the first two are equivalent.

Suppose $f$ satisfies the 2nd definition above, and pick $m$ as given by the definition. Let $q_x(\Delta x) = m$, which does not depend on $\Delta x$ and hence has a limit as $\Delta x\to 0$). Let $r_x(\Delta x) = f(x+\Delta x)-f(x) - m\Delta x$. We must prove that $r_x(\Delta x)/ \Delta x\to 0$ as $\Delta x\to 0$. Let $\varepsilon>0$. Pick $\delta>0$ such that for every $\Delta x$ with $x+\Delta x\in U$, we have if $0 < |\Delta x|<\delta$ then $|f(x+\Delta x)-f(x) - m\Delta x| < |\Delta x| \varepsilon$. This means that $|\frac{f(x+\Delta x)-f(x) - m\Delta x}{|\Delta x}|-0| < \varepsilon$, which is equivalent to This means that $|\frac{r_x(\Delta x)}{|\Delta x}|-0| < \varepsilon$.

Assume now that (4) is true. We will prove that (2) is true. Pick $q_x(\Delta)$ and $r_x(\Delta x)$ such that

  • $q_x(\Delta x)$ has a limit as $\Delta x\to 0$,
  • $r_x(\Delta x)/|\Delta x|\to 0$ as $\Delta x\to 0$, and
  • for every $\Delta x\neq 0$ with $x+\Delta x\in U$, we have $f(x+\Delta x)-f(x) = q_x(\Delta x)\Delta x +r_x(\Delta x)$.

Let $m = \lim_{\Delta x\to 0}q_x(\Delta x)$. We must prove for every $\varepsilon>0$, there exists a $\delta >0$ such that for every $\Delta x$ with $x+\Delta x\in U$, we have if $0 < |\Delta x|<\delta$ then $|f(x+\Delta x)-f(x) - m\Delta x| < |\Delta x| \varepsilon$. Note that $$\begin{align} |f(x+\Delta x)-f(x) - m\Delta x| &= |q_x(\Delta x)\Delta x +r_x(\Delta x) - m(\Delta x)| \\ &= |q_x(\Delta x)\Delta x- m(\Delta x) +r_x(\Delta x) | \\ &= |(q_x(\Delta x)-m)(\Delta x)+r_x(\Delta x) | (\text{sum of linear operators is linear})\\ &<= |(q_x(\Delta x)-m)| |\Delta x|+|r_x(\Delta x) | (|Ax|\leq |A||x| \text{ and triangle ineq} \\ &=|(q_x(\Delta x)-m)| |\Delta x|+|\frac{r_x(\Delta x) }{|\Delta x|}-0| |Delta x| \\ \end{align}$$ Let $\varepsilon>0$. Pick $\delta_1>0$ such that $|(q_x(\Delta x)-m)|<\varepsilon/2$. Pick $\delta_2>0$ such that $|\frac{r_x(\Delta x) }{|\Delta x|} - 0|<\varepsilon/2$. Done.

The above requires way more information than I want to include. Can I simplify it in 1D nicely. Here is what I want

We say that $f$ is differentiable at an interior point $x\in (a,b)$ if we can write $$f(x+dx)-f(x) = m_{x,dx} dx +r_{x,dx} dx$$ for nonzero $dx$ close to $x$ such that $\lim_{dx\to 0} m_dx$ exists and $r_{x,dx} \to 0$ as $dx \to 0$.