3
$\begingroup$

I can't figure out the intuition and motivation behind the following definition:

Function $f$, defined in some neighborhood of $x_0$, is said to be differentiable, if: $$f(x_0 + \Delta x)-f(x_0)=A\Delta x+o(\Delta x), \Delta x \rightarrow 0$$ where $A$ is some real number, or equivalently: $$\Delta y=A\Delta x +\epsilon(\Delta x)\Delta x$$ where $\epsilon(\Delta x)$ is infinitely small as $\Delta x \rightarrow 0$.

I understand the limit definition of differentiability, and I understand the proof that two definitions are equivalent, but I just don't get the above one. At first look it says that we can approximate the change of function by some linear(?) function ? And if it's saying that we can approximate the value of $f$ at point $a$ by some linear function, doesn't this definition make every function differentiable, as we can chose constant function $f(x)=a$ for approximation .. ? Also, I don't get the $o(\Delta x)$ part .. Can this be shown visually ? (But then, visualization would be the same as for limit definition, right?)

$\endgroup$
3
  • 1
    $\begingroup$ The idea of derivatives is that a differentiable function can be locally approximated by a linear function, and we like linear functions since we can model them with our linear tools. $\endgroup$ Mar 30, 2019 at 15:28
  • $\begingroup$ I understand the idea of derivatives and the algebra in the proof that two definitions are equivalent, but I don't understand intuitively that two definitions are stating the same thing. $\endgroup$
    – user655080
    Mar 30, 2019 at 15:32
  • $\begingroup$ Do you get it if the $o(\Delta x)$ isn't there? $\endgroup$ Mar 30, 2019 at 15:46

1 Answer 1

2
$\begingroup$

If you understand the definition of derivative then you know that $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

This means that for small $h$ $$f'(x)\approx\frac{f(x+h)-f(x)}{h}$$

More precisely, there is an error term $\epsilon$ such that $$f'(x)=\frac{f(x+h)-f(x)}{h}+\epsilon$$ Note that, in this case, $\epsilon$ doesn't have to be positive.

Multiplying by $h$ gives us $hf'(x)=f(x+h)-f(x)+\epsilon h$ and subtracting $\epsilon h$ from both sides yields $$f(x+h)-f(x)=hf'(x)-\epsilon h$$

Now, replace $-\epsilon h$ by $+\epsilon h$ and $h$ by $\Delta x$ to get the equivalent definition: $$f(x+\Delta x)=f'(x)\Delta x+\epsilon\Delta x$$ where $\epsilon\to 0$ as $\Delta x\to 0$. Divide through by $\Delta x$ and take the limit to see why $\epsilon$ must go to $0$ as $\Delta x$ goes to $0$.

I hope you can see that there is no intuition behind this besides the fact that we have rewritten the derivative.

We do this because we want to be able to define the derivative of functions on higher dimensional spaces.

For example, suppose $f:\mathbb R^2\to\mathbb R$ is defined by $f(x,y)=x+y$. Is it not valid to ask how $f$ changes if its inputs are changed? We can't use our original definition because vector division is not defined.

We can say that $f$ is differential if there exists a function $A:\mathbb R^2\to\mathbb R$ and an error term $\epsilon(\vec{h})$ such that $$f(\vec{x} +\vec{h})=A\vec{h}+\epsilon(\vec{h})$$

where $\lim_{|\vec{h}|\to 0}\epsilon=0$

$\endgroup$
2
  • $\begingroup$ "I hope you can see that there is no intuition behind this besides the fact that we have rewritten the derivative" - this, thank you! I was really struggling with figuring out why is there a need to define the derivative in such way (I think I get the use of it for more than one variable functions). $\endgroup$
    – user655080
    Mar 30, 2019 at 16:42
  • $\begingroup$ @ubermensch You are welcome and if you have been shown this in a class you will see that this is where your teacher is going. $\endgroup$
    – John Douma
    Mar 30, 2019 at 16:56

You must log in to answer this question.