(Full disclosure: I'm a complete amateur at differentiation, and just trying to understand derivatives through this question)
My explanation of derivatives:
Consider the curve $y=x^3$:
(Let us only consider the 1st quadrant for now for simplicity)
Now, what is the slope of this curve? Answer: it doesn't have a fixed slope like that of a straight line; its slope is constantly changing. Now, a better question is, where exactly does the slope change? I'll attempt to answer this question now:
Consider this graph:
Now, the graph of the above function does not have a fixed slope. It has 3 well-defined slopes in the regions OA, AB & BC. Now, where exactly do the slopes of the function change? Answer: it changes 2 times: first, at point A before the start of segment A, and second at point B before the start of segment BC. Okay, good.
Now, what if we made the regions OA, AB & BC smaller. Okay, that's fine; however, it won't be the same function anymore:
Now, there are 6 regions with well-defined slopes, OA, AB, BC, CD, DE & EF, and the slope changes 5 times at the points A, B, C, D & E.
Now, if we keep making the regions smaller and smaller, we will get the function that we started with, $y=x^3$:
Now, back to our original question, where exactly does the slope of the graph change? The answer is, at each of the infinitely many points of the graph! You can visualize this in this way: in our previous examples, there were a finite number of points where the slope changed. However, if we keep making the regions smaller and smaller, the number of regions and the number of times the slope changes will get bigger and bigger, and at one point, the regions will lose their 2-dimensionality: it will become one-dimensional, just a point, and we will get $y=x^3$. But the slopes are still changing, so we can understand that the slopes are changing at every point of the curve! Where else is the slope changing?!
Now, you might object that how can a point have a slope? You need two points to calculate a slope; you only have one point. Answer: you are right. A point in isolation can't have any slope; however, a point that is part of a graph can have a well-defined slope. Get this, if points can't have slopes, how is the slope of a curve changing at every point? The fact that the slope of the curve is changing at every point is proof that a point can have a slope. Okay, now I will define what the slope of a point is.
The slope of a point that is part of the graph of a function is the slope of the line that is tangent to the graph of the function at that point…(i)
Now, how do we calculate this slope of the tangent line? Luckily, earlier mathematicians have worked this out for us:
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
$f'(x)$ is the slope of the line tangent to the graph of $f$ at $x$. $f'(x)$ is also known commonly as the derivative of $f$ at $x$: this is just another name for the slope of the tangent line.
Questions:
- Is (i) a legitimate definition in mathematics?
- Is my explanation correct? Do you disagree with any parts of my explanation?
Bonus question:
(If you don't want to answer this, it's fine! However, if you do, it might be helpful for me.)
- Was my thinking process similar to that of the fathers of calculus? If not, what was their thinking process?
Best Answer
Your reasoning is essentially correct and capture the intuitive geometrical meaning of the derivative. But a rigorous definition is more difficult. Your definition (i) can work only if we have a previous rigorous definition of what is a"tangent" to a curved line. But such a rigorous definition requires the definition of the "slope" at the point, so the problem seems to be a vicious circle from which we can escape using the concept of limit.
What we means exactly for a "straight line tangent to a point"? We know that a straight line is well defined when we have two points, so, given a point $(x,f(x))$ and a point $(x+h,f(x+h))$ we can write the slope of the function that passes between these points as: $$ \frac{f(x+h)-f(x)}{h} $$ than we define the tangent as the line that passes through $(x,f(x))$ and has slope the limit of this quotient when $h \rightarrow 0$ if this limit exists.
So, at the same time, we define the tangent and his slope, that can be thinked as the slope of the function at the given point.