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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 2

Lesson 6: Derivative rules: constant, sum, difference, and constant multiple: introduction# Basic derivative rules: find the error

AP.CALC:

FUN‑3 (EU)

, FUN‑3.A (LO)

, FUN‑3.A.2 (EK)

Sal analyzes two attempts of students to differentiate linear functions.

## Want to join the conversation?

- why the derivative the same as the differentiation.(0 votes)
- Differentiation
, it is what you do to calculate a derivative.**is a process**

The derivative, it is the**is a function***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(33 votes)

- 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?(9 votes)
- 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.(3 votes)

- How does d(x)/dx turn out to be 1?(3 votes)
- It's the rate of change, or slope, of the function f(x)=x, which is a straight line of slope 1.(5 votes)

- 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?(2 votes)
- 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!(5 votes)

- at2:25, 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(2 votes)- 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.(3 votes)

- Why Hannah did is wrong?(2 votes)
- 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.(2 votes)

- 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!(1 vote)

## 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.