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## Differential Calculus (2017 edition)

### Unit 5: Lesson 6

Formal definition of derivative

# Worked example: Derivative from limit expression

AP.CALC:
CHA‑2 (EU)
,
CHA‑2.B (LO)
,
CHA‑2.B.2 (EK)
,
CHA‑2.B.3 (EK)
,
CHA‑2.B.4 (EK)
Sal interprets a limit expression to find that it describes the derivative of f(x)=x³ at the point x=5. Created by Sal Khan.

## Want to join the conversation?

• It would be nice if the video focused more on finding the derivative at an indicated point or a separate video for it. I'm great at the Formal form but the alternate form is more difficult for me. Even just a simple f(x)= 1/x, a=2 explanation would be great. If someone could help me out that would be great as well.
• If I understand you're question correctly, you'd like to have an explanation of how to use alternate form to get the derivative at a point. Hopefully my explanation below isn't too long >n<. I also tried to explain how formal and alternate form are different ways of saying the same thing.

Let's do this example you have provided, f(x)=1/x, with a=2.

For the sake of demonstration, let's do the formal form first. Formal form is saying "I'm going to find the derivative by decreasing a distance from my point at 2 down to 0".

You would probably write it out like this:
f'(x)=lim (h->0) ((1/(2+h)-1/2)/(h)
I'll let you crack the algebra to save space, but unless I made an error you should eventually boil it down to this:
-1/2(2+h)
and as h approaches 0, you get -1/4.

Ok, now let's talk alternate form. Instead of taking a point h distance away from 2 and reducing that distance to 0, I'm going to say it another way- "I'm going to just take a point x, and bring it closer to 2." If you think about it, it's the same thing. Once x approaches 2, the distance between the two will be 0-just like with the distance h from 2 in formal form. It's just an alternate way of finding the same thing. To put it more succinctly, we're substituting h with x, and just saying that the closer we get x to 2 is the same thing as saying we're moving a point 2+h closer to 2 by reducing h to 0.

So let's write it out:
f'(x)=lim (x->a) (1/x-1/2)/(x-2)

If you do the algebra (hint: Get the denominators at the top to play nice first), you'll eventually get:

-1/2x

Plug in 2 for x (remember, we're having x approach 2), and you'll get the same thing as formal form- -1/4.

Hope that was what you were looking for and it wasn't too long. Let me know if this helps.
• Any tip to understand the difference(s) between the formal form and the alternate one clearer and to apply them better? In what case(s) we should use the prior, and in what case(s) the latter is better?
• Generally, we will use the formal form of a derivative when trying to find elementary derivatives like finding the derivative of y = x, or the derivative of y = x^2. The alternate form is still useful, although it is primarily used in finding slopes of lines, not curves.
Differences between formal form and alternate form are in how the limits are defined. In both, we are letting two points approach each other, but one uses lim h --> 0 and the other uses lim x --> a.
• I calculated the slope of the tangent line of f(x)=x^3 at x=5 to be 75. Can anyone confirm this for me?
• That is correct. The derivative of x^3 is 3x^2, and if you plug in x=5, that comes out to 75.
• The secant line is found by (Y2-Y1 / X2-X1) I always thought the (X2 , Y2) were the point furthest to the right? But in this video () he is coming from the left. How do we know what values to put in ( f(x)-f(a) / x-a ) ? or does it matter what order they are in?
• The secant line is basically the slope of the 2 points. In other words, changes in Y over changes in X. It doesn't matter from which direction you approach it from as long as you are consistent with your order pair, meaning, let (x,f(x)) and (a,f(a)) be the two points, you can use f(x)-f(a)/x-a or f(a)-f(x)/a-x.

Generally, in the book it will teach you that f(x) is on the order pair from the right and > f(a) this is because it's more intuitive when doing the subtraction. But just remember that the secant line is just the slope between any two points on the graph. And if you know how to find the slope of any two points, then you can find the secant line.
• Knowing only the formal and the alternative form of the derivative, are we supposed to know how to evaluate f'(x) or f'(a)? Specifically, I am having troubles simplifying (x^3 - 125)/(x-5) so that I can take the limit of it as x approaches 5. If so, I might have just missed something and I will go back some videos to revise.
• what is a prime function with the symbol ' such as f'(a) ?
(1 vote)
• f'(a) denotes the first derivative of f(a)
• How do you find f(x)= 14x+13, if f'(1) or f prime of 1
• I'm not sure I completely understand your question, but since your function is linear, the slope is constant, so the derivative is also constant. Your equation is of the linear form: y=mx+b where m is the slope. In your case it is 14, so the derivative is f'(x)=14.
f'(1)=14 too
• I thought the derivative was going to be an actual value. Why was it kept in that form?
(1 vote)
• Normally if a function is differentiable, then you'd want to write down what that resulting function is so that you can compute it at any point you'd like - instead of a numerical value. For instance if f(x) = x^2 then f'(x) = 2x. This is true for any value of x.
• I am needing help with a problem.

Use the 4 step process to find f'(x) and then find f'(1) f'(2) f'(3)

f(x)=3x-7
(1 vote)
• 𝑓(𝑥) = 3𝑥 − 7

Step 1: 𝑓(𝑥 + ℎ) = 3(𝑥 + ℎ) − 7 = 3𝑥 − 7 + 3ℎ

Step 2: 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 3𝑥 − 7 + 3ℎ − (3𝑥 − 7) = 3ℎ

Step 3: (𝑓(𝑥 + ℎ) − 𝑓(𝑥))∕ℎ = 3ℎ∕ℎ = 3

Step 4: lim(ℎ → 0) (𝑓(𝑥 + ℎ) − 𝑓(𝑥))∕ℎ = lim(ℎ → 0) 3 = 3

So, 𝑓 '(𝑥) = 3 ⇒ 𝑓 '(1) = 𝑓 '(2) = 𝑓 '(3) = 3
• Why is the derivative of all these functions just algebraic expressions? Don't they have any numerical value? Because I remember in the intro of calculus that we can use it to find the instantaneous value of a curve. If they do have a value, how can we find it? For e.g., what is the numerical value of f'(5) in the above problem?
(1 vote)
• Because you are solving for the general derivative of the functions.To find the particular solution for a X-value, all you have to do is plug in the X-value into the derivative.

For your example of f'(5), as f(x) = x^3
f'(x) = 3x^2
So you plug in 5 for the X-value
f'(5) = 3(5)^2 = 75

Also, this video solves for the derivative using the formal definition but there is another way that you will learn later which is much faster and easier (in my opinion).