Explanation of the derivative as a function

[jpanknin]

Trying to wrap my head around why a derivate function is always right. For example, [imath]\frac{d}{dx} x^2 = 2x[/imath] and [imath]\frac{d}{dx} x^3 = 3x^2[/imath], where [imath]f(x) = x^2 == > f'(x) = 2x[/imath] and [imath]g(x) = x^3 == > g'(x) = 3x^2[/imath]. These derivative functions, [imath]f'(x)[/imath] and [imath]g'(x)[/imath], will give the right slope regardless of whether x is 3, 412, -19, or any other real number. I've worked it out using definitions of the derivative and they obviously work, but it seems a bit "magical" that they ALWAYS give the right slope regardless of the value of x. Is there a better/different way to think about what's happening here?

 

[topsquark]

Trying to wrap my head around why a derivate function is always right. For example, [imath]\frac{d}{dx} x^2 = 2x[/imath] and [imath]\frac{d}{dx} x^3 = 3x^2[/imath], where [imath]f(x) = x^2 == > f'(x) = 2x[/imath] and [imath]g(x) = x^3 == > g'(x) = 3x^2[/imath]. These derivative functions, [imath]f'(x)[/imath] and [imath]g'(x)[/imath], will give the right slope regardless of whether x is 3, 412, -19, or any other real number. I've worked it out using definitions of the derivative and they obviously work, but it seems a bit "magical" that they ALWAYS give the right slope regardless of the value of x. Is there a better/different way to think about what's happening here?

I really don't understand your question. If you have derived the power rule via Calculus, then that's what the derivative is.

Somewhat more formally, define the space of continuous real functions on x to be [imath]C^0( \mathbb{R} )[/imath]. Then
[imath]\dfrac{d}{dx} : C^0( \mathbb{R} ) \to C^0( \mathbb{R} )[/imath]

such that
[imath]\displaystyle \dfrac{d}{dx} f(x) \equiv \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}[/imath]

(Actually, the target space isn't [imath]C^0 (\mathbb{R})[/imath]: the derivative may not be continuous. But anyway...)

This is a linear map.

Is this what you are going for? Or am I barking up the wrong tree?

-Dan

 

[lev888]

Trying to wrap my head around why a derivate function is always right. For example, [imath]\frac{d}{dx} x^2 = 2x[/imath] and [imath]\frac{d}{dx} x^3 = 3x^2[/imath], where [imath]f(x) = x^2 == > f'(x) = 2x[/imath] and [imath]g(x) = x^3 == > g'(x) = 3x^2[/imath]. These derivative functions, [imath]f'(x)[/imath] and [imath]g'(x)[/imath], will give the right slope regardless of whether x is 3, 412, -19, or any other real number. I've worked it out using definitions of the derivative and they obviously work, but it seems a bit "magical" that they ALWAYS give the right slope regardless of the value of x. Is there a better/different way to think about what's happening here?

Please post the definition of a derivative function.

 

[skeeter]

… but it seems a bit "magical" that they ALWAYS give the right slope regardless of the value of x.

 

[Dr.Peterson]

Trying to wrap my head around why a derivate function is always right. For example, [imath]\frac{d}{dx} x^2 = 2x[/imath] and [imath]\frac{d}{dx} x^3 = 3x^2[/imath], where [imath]f(x) = x^2 == > f'(x) = 2x[/imath] and [imath]g(x) = x^3 == > g'(x) = 3x^2[/imath]. These derivative functions, [imath]f'(x)[/imath] and [imath]g'(x)[/imath], will give the right slope regardless of whether x is 3, 412, -19, or any other real number. I've worked it out using definitions of the derivative and they obviously work, but it seems a bit "magical" that they ALWAYS give the right slope regardless of the value of x. Is there a better/different way to think about what's happening here?

I think you're saying that you understand the calculation of the derivative at a particular point, but don't grasp the idea of f' as a new function that gives the slope of the graph of f at any point. Am I right? This does often take some time to absorb.

One way to think about it is just to imagine finding the slope of a given curve at some x=a and plotting the point (a, f'(a)), so that the y-coordinate of this point is the slope of the original function. If you could plot all these points, you would have a new curve; that curve is the graph of the function f'. That's all it is.

This textbook section may help a little:

3.2: The Derivative as a Function

The derivative of a function f(x) is the function whose value at x is f′(x). The graph of a derivative of a function f(x) is related to the graph of f(x). Where (f(x) has a tangent line with …

math.libretexts.org

But you may need to explain your thinking a little more specifically, maybe with examples, so we can answer your particular thinking.

 

[jpanknin]

I think you're saying that you understand the calculation of the derivative at a particular point, but don't grasp the idea of f' as a new function that gives the slope of the graph of f at any point. Am I right? This does often take some time to absorb.

This right here is what the problem is. I have no problem calculating it, but to reference the cartoon @skeeter posted, it does seem a bit miraculous that the calculation will extend to the slope of the tangent at ANY point on the graph of f, regardless of the value of x or the direction of the graph. Again, I have no problem with the math behind it, just having trouble seeing the jump from the graph of f(x) to f'(x) intuitively.

One way I've thought of it is that if we let f'(x) = g(x) then it essentially takes the concept of a derivative out of the problem and just gives the expression of a different function, which you can then graph the same way you graphed f(x). I think this is similar to what you've described in the second paragraph, @Dr.Peterson

 

[lev888]

just having trouble seeing the jump from the graph of f(x) to f'(x) intuitively.

There is a reason I asked you to post the definition. Do you understand the definition? Are you familiar with limits? Do you understand limits intuitively?

 

[jpanknin]

@lev888 Yes, I understand the limits and the definitions (see below). It makes more intuitive sense to me when we're using real numbers and can calculate the slope (derivative) at specific points, say [imath]f(x) = x^2[/imath] and then plug in various values for h that approach 0. It's taken me 20 minutes to write this because I kept trying different calculations to get my thinking straight. The definitions below make intuitive sense to me. No matter what value for x you use and then add an arbitrarily small value of h to x and then take the limit, you get the derivative. I tried several problems and then graphed the resulting tangent line and it works. Again, I think I understand these definitions and the math behind them pretty well. It's the jump from these definitions to the magical f'(x) = 2x that works everywhere for every number that still seems like magic.

 

[Dr.Peterson]

This right here is what the problem is. I have no problem calculating it, but to reference the cartoon @skeeter posted, it does seem a bit miraculous that the calculation will extend to the slope of the tangent at ANY point on the graph of f, regardless of the value of x or the direction of the graph. Again, I have no problem with the math behind it, just having trouble seeing the jump from the graph of f(x) to f'(x) intuitively.

One way I've thought of it is that if we let f'(x) = g(x) then it essentially takes the concept of a derivative out of the problem and just gives the expression of a different function, which you can then graph the same way you graphed f(x). I think this is similar to what you've described in the second paragraph, @Dr.Peterson

I think you've got the idea. I don't think it's intuitive at all; you have to train your intuition to accept it! We aren't used to deriving one function from another (though that is exactly where the word "derivative" comes from).

But if you think about it, a graph is the same idea: a single picture that describes something you do to x, for all possible values of x, all at once. The "something" in this case is just finding the slope of f.

So in principle it shouldn't be hard. We have something we can do at any point on a curve to find its slope, and we choose to call the resulting number a function of that point on the curve. If the limit definition isn't magical when applied at one point, then leaving x as a variable and doing the same limit for every possible point at once ... is just what algebra does all the time. We just don't usually do that until we define the derivative, so it's an unfamiliar perspective.

One way to relate it to everyday life is to think of the speed of a car, which is the derivative of its position. If you recorded the speed, you would get a graph which at each point in time shows the rate of change of the car's position. You might try looking at Example 3.2.3 in the lesson I linked to with this in mind, thinking of a car's position as given by f(t).

 

[lev888]

@lev888 Yes, I understand the limits and the definitions (see below). It makes more intuitive sense to me when we're using real numbers and can calculate the slope (derivative) at specific points, say [imath]f(x) = x^2[/imath] and then plug in various values for h that approach 0. It's taken me 20 minutes to write this because I kept trying different calculations to get my thinking straight. The definitions below make intuitive sense to me. No matter what value for x you use and then add an arbitrarily small value of h to x and then take the limit, you get the derivative. I tried several problems and then graphed the resulting tangent line and it works. Again, I think I understand these definitions and the math behind them pretty well. It's the jump from these definitions to the magical f'(x) = 2x that works everywhere for every number that still seems like magic.

View attachment 36171

Given the definition of a function, we know its value for any number of its domain. And we obviously know exactly how it behaves in the neighborhood of each number. The definition of a derivative function makes sense - for any point x it's just the slope at point x. But at the same time, it's a function of x. So the definition is a method for constructing the derivative function. We apply this method to several function types, function composition, etc and, as a result, we can find derivatives for any function that can be differentiated.

Consider this, much simpler, example. You graph a function f(x). Then you ask yourself: "I wonder what it would look like if I added 10 to each y value." What do you do? You add 10 to the expression of the function, simplify it, and graph the new function. Its points are exactly 10 units higher than the original points. Do you consider the new function "magical"?

 

[Steven G]

Although this particular issue never bothered me, other math concepts have. The answer to your question is that it follows from the definition of derivatives which you claim to understand. OK, think it magical--I don't (or neither should you) have a problem with that. The thing is it can be proven that the derivative of sums/differences is the sums/differences of derivatives and the power rule can be proven. That's it! It works because we proved it to work. It's cool, magical, amazing,....--but it's true.
Welcome to the world of pure mathematics

 

[Steven G]

How long have you been waiting for a chance to post this cartoon? I think that you posted it before.

 

[skeeter]

How long have you been waiting for a chance to post this cartoon? I think that you posted it before.

Probably have posted it before ... posted it in this thread because it seemed an appropriate response to the "magical" reference.

Why do you ask?

 

[Steven G]

Probably have posted it before ... posted it in this thread because it seemed an appropriate response to the "magical" reference.

Why do you ask?

I think that it is cool that you wait for the right moment to post that cartoon.