For a linear equation in the form y=mx+b, the coefficient of the x term is the slope of the line. Since the derivative of a function is the slope at any given point and since the slope of a line is always constant, is the derivative not always equal to the coefficient of the x term? Why do we take all of these additional steps?

Is to proof the basic differentiantion rules. Lineal functions are intuitive, but when you are dealing with more complex functions like x^3+x^2+e^x-log_2(x), they become really usefull.

How does d(x)/dx turn out to be 1?

It's the rate of change, or slope, of the function f(x)=x, which is a straight line of slope 1.

I thought derivatives were for particular points of a function, not whole functions themselves. Why didn't he choose a certain x-value of each function to differentiate?

That's the cool thing about lines! The derivative of any linear function is a _constant_, meaning no matter what 𝑥-value you choose, the derivative is always the same. For instance, the derivative of 𝑓(𝑥) = 5𝑥 is 𝑓'(𝑥) = 5. This is 5 no matter what 𝑥 is! Informally, we say that the slope of a line is _constant_ everywhere. Comment if you have questions!

at 2:25, derivative of 5*f(x) for any x, x is an actual point which means is a constant right? then why sal says is 1? shouldn't derivative of any constant going to be 0? i understand graphicly slop of y=x is going to be 1 but not algebraicly

To find a derivative at a point, we don't plug in the point, solve, then take the derivative and get 0. In doing so, we assume that the function is equal to that constant value everywhere, which is untrue (in general). The derivative of the function y=x at the point a is the limit as a→x of (a-x)/(a-x), which is the limit of the constant 1, which is 1.

Why Hannah did is wrong?

Let 𝑓(𝑥) = 8 and 𝑔(𝑥) = 𝑥 ⇒𝑓 '(𝑥) = 0 and 𝑔'(𝑥) = 1 Then 𝑑∕𝑑𝑥[8𝑥] = 𝑑∕𝑑𝑥[𝑓(𝑥)⋅𝑔(𝑥)] Applying the product rule, we get 𝑓 '(𝑥)⋅𝑔(𝑥) + 𝑓(𝑥)⋅𝑔'(𝑥) This is not equivalent to 𝑓 '(𝑥)⋅𝑔'(𝑥) which is what Hannah did.

why the derivative the same as the differentiation.

Differentiation _*is a process*_, it is what you do to calculate a derivative. The derivative _*is a function*_, it is the _result of differentiation_. EG f(x)=x² - - - this is the original function d/dx[f(x)=x²] - - - this is the original function inside the differentiation operator f'(x)=2x - - - this function is the derivative of the original function. We calculated it using the process of differentiation. So . . . d/dx[f(x)=x²] - - - differentiation f'(x)=2x - - - derivative

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Calculus, all content (2017 edition)

Course: Calculus, all content (2017 edition) > Unit 2

Lesson 11: Basic differentiation rules

Basic derivative rules: find the error

Name: Basic derivative rules: find the error
Uploaded: 2016-08-18T15:48:49Z
Description: We analyze two students' attempts to differentiate linear functions using basic rules. One student forgets a negative sign, while another incorrectly assumes the derivative of a product equals the product of derivatives. By identifying common mistakes, we help improve our understanding of differentiation.

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We analyze two students' attempts to differentiate linear functions using basic rules. One student forgets a negative sign, while another incorrectly assumes the derivative of a product equals the product of derivatives. By identifying common mistakes, we help improve our understanding of differentiation.

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cherdchai
Posted 7 years ago. Direct link to cherdchai's post “why the derivative the sa...”
why the derivative the same as the differentiation.
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- Stefen
  Posted 7 years ago. Direct link to Stefen's post “Differentiation _*is a pr...”
  Differentiation is a process, it is what you do to calculate a derivative.
  The derivative is a function, it is the result of differentiation.
  EG
  f(x)=x² - - - this is the original function
  d/dx[f(x)=x²] - - - this is the original function inside the differentiation operator
  f'(x)=2x - - - this function is the derivative of the original function. We calculated it using the process of differentiation.
  So . . .
  d/dx[f(x)=x²] - - - differentiation
  f'(x)=2x - - - derivative
  Comment on Stefen's post “Differentiation _*is a pr...”
  (37 votes)
david parbel
Posted 7 years ago. Direct link to david parbel's post “For a linear equation in ...”
For a linear equation in the form y=mx+b, the coefficient of the x term is the slope of the line. Since the derivative of a function is the slope at any given point and since the slope of a line is always constant, is the derivative not always equal to the coefficient of the x term? Why do we take all of these additional steps?
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(10 votes)
Answer
- Juan Eduardo Ynsil Alfaro
  Posted 7 years ago. Direct link to Juan Eduardo Ynsil Alfaro's post “Is to proof the basic dif...”
  Is to proof the basic differentiantion rules. Lineal functions are intuitive, but when you are dealing with more complex functions like x^3+x^2+e^x-log_2(x), they become really usefull.
  Button navigates to signup page
  (3 votes)
Jyotiraditya Pradhan
Posted 3 years ago. Direct link to Jyotiraditya Pradhan's post “How does d(x)/dx turn out...”
How does d(x)/dx turn out to be 1?
Button navigates to signup pageComment on Jyotiraditya Pradhan's post “How does d(x)/dx turn out...”
(3 votes)
Answer
- kubleeka
  Posted 3 years ago. Direct link to kubleeka's post “It's the rate of change, ...”
  It's the rate of change, or slope, of the function f(x)=x, which is a straight line of slope 1.
  Comment on kubleeka's post “It's the rate of change, ...”
  (5 votes)
20leunge
Posted 6 years ago. Direct link to 20leunge's post “I thought derivatives wer...”
I thought derivatives were for particular points of a function, not whole functions themselves. Why didn't he choose a certain x-value of each function to differentiate?
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(2 votes)
Answer
- andrewp18
  Posted 6 years ago. Direct link to andrewp18's post “That's the cool thing abo...”
  That's the cool thing about lines! The derivative of any linear function is a constant, meaning no matter what 𝑥-value you choose, the derivative is always the same. For instance, the derivative of 𝑓(𝑥) = 5𝑥 is 𝑓'(𝑥) = 5. This is 5 no matter what 𝑥 is! Informally, we say that the slope of a line is constant everywhere. Comment if you have questions!
  Button navigates to signup page
  (6 votes)
Ethan.M
Posted 5 years ago. Direct link to Ethan.M's post “at 2:25, derivative of 5*...”
at
2:25
, derivative of 5*f(x) for any x, x is an actual point which means is a constant right?
then why sal says is 1? shouldn't derivative of any constant going to be 0?
i understand graphicly slop of y=x is going to be 1 but not algebraicly
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(2 votes)
Answer
- kubleeka
  Posted 5 years ago. Direct link to kubleeka's post “To find a derivative at a...”
  To find a derivative at a point, we don't plug in the point, solve, then take the derivative and get 0. In doing so, we assume that the function is equal to that constant value everywhere, which is untrue (in general).
  
  The derivative of the function y=x at the point a is the limit as a→x of (a-x)/(a-x), which is the limit of the constant 1, which is 1.
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  (3 votes)
chenchen
Posted 2 years ago. Direct link to chenchen's post “Why Hannah did is wrong?”
Why Hannah did is wrong?
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(2 votes)
Answer
- Jerry Nilsson
  Posted 2 years ago. Direct link to Jerry Nilsson's post “Let 𝑓(𝑥) = 8 and 𝑔(𝑥)...”
  Let 𝑓(𝑥) = 8 and 𝑔(𝑥) = 𝑥
  ⇒𝑓 '(𝑥) = 0 and 𝑔'(𝑥) = 1
  
  Then 𝑑∕𝑑𝑥[8𝑥] = 𝑑∕𝑑𝑥[𝑓(𝑥)⋅𝑔(𝑥)]
  
  Applying the product rule, we get
  𝑓 '(𝑥)⋅𝑔(𝑥) + 𝑓(𝑥)⋅𝑔'(𝑥)
  
  This is not equivalent to
  𝑓 '(𝑥)⋅𝑔'(𝑥)
  which is what Hannah did.
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  (2 votes)
Garret Cervantez
Posted 6 years ago. Direct link to Garret Cervantez's post “This is a little differen...”
This is a little different notation so tell me if I understand this correctly.
f'(x) means what is the derivative for the function f(x) at x.
d/dx(x) means what is the derivative for the function f(x) = x.

Am I understanding this correctly? Thanks!
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Answer

Video transcript

- [Voiceover] So we have two examples here of someone trying to find the derivative of an expression. On the left-hand side, it says "Avery tried to find the derivative, "of seven minus five x using basic differentiation rules. "Here is her work," and on the right-hand side it says "Hannah tried to find the derivative, "of negative three plus eight x, "using basic differentiation rules, "here is her work." And these are two different examples of differentiation rules exercise on Khan academy, and I thought I would just do them side by side, because we can kind of think about what each of these people are doing correct or incorrect. So these are similar expressions, we have a constant and then we have a first degree term, a constant and then first degree term. So they're gonna take the derivative, so let's see, step one for Avery. She took, she's separately taking the derivative of seven, and separately taking the derivative of five x. So this my spider senses already going off here, because what happened to this negative right over here? So it would've sense for her to do the derivative of seven, and she could've said minus the derivative, of five x, that's one possibility that she could've done. The derivative of a difference is equal to the difference of the derivatives, we've seen that property. Or, she could've said, the derivative, she could've said this was equal to the derivative of seven, plus the derivative with respect to x of negative five x. These two things would've been equivalent to this one. But for this one, she somehow forgot to, include the negative. So I think she had a problem right at step one. Now, if you just follow her logic after step one, let's see if she makes any more mistakes. So, she takes the derivative of a constant, so constant isn't going to change with respect to x. So that makes sense, that that derivative is zero. And so we still have the derivative of five x, and remember, it should've been negative five x, or minus the derivative of five x. And let's see what she does here. So that zero disappears, and now she takes the constant, she takes the constant out, and that's true, the derivative of a constant times something, is equal to the constant times the derivative of that something. And then, she finds the derivative with respect to x of x is one, and that's true, if the slope, if you had the graph of y equals x, the slope there is one, or what's the rate of change at which x changes with respect to x? Well, that's gonna be one for one. So that the slope here is one, so this is gonna be five times one. Which is equal to five, and at the end they just say, at what step did Avery make a mistake? So she clearly made a mistake at step one, this right of here, should've been a negative, that's a negative, then that would've been a negative. And this would've been a negative. And that would've been a negative. And then her final answer should have been, should have been a negative five. Now let's go back to Hannah. To see if she made any mistakes and where. So she's differentiating a similar expression, so first she takes the derivative of the constant, plus the derivative of the first degree term. Derivative of constant is zero, that looks good. So you get the zero, and then you have the derivative of the first degree term. That's what she's trying to figure out. And then, let's see, she's taking... Let's see, so this seems off. She is assuming that the derivative of a product is equal to the product of the derivatives. That is not the case. And especially, and if you have a constant here, there's actually a much simpler way of thinking about it. Frankly the way that Avery thought about it, Avery had made a mistake at step one, but this is actually going to be equal to the derivative of a constant times an expression, is equal to the same thing as the constant times the derivative of, of the expression. So this would've been the correct way to go, and the derivative of x with respect to x, well that's just going to be one. So this should've all simplified to eight. What she did is, she is assuming, she tried to take the derivative of eight and multiply that times the derivative of x, that is not the way it works. In the future you will learn something called the product rule, but you won't even have to apply that here, because one of these, one of these components I guess you could say, is a constant. So this is the wrong step. This is where Hannah makes a mistake. And you could see, instead of getting a final answer of eight, she is getting a final answer of, she assumes well the derivative of eight is zero, times the derivative of x is one, zero times one. And she gets zero, which is not the right answer. So she makes a mistake at step three, and Avery made a mistake at step one.