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## Calculus, all content (2017 edition)

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

Lesson 11: Basic differentiation rules

# Proof of the constant derivative rule

Sal introduces the Constant rule, which says that the derivative of f(x)=k (for any constant k) is f'(x)=0. He also justifies this rule algebraically.

## Want to join the conversation?

• Is it really justifiable this way? The lim h>0 (k-k) / h equals 0/0. We are not allowed to divide by zero.

Certainly, L'Hopital rule may be possible to apply here but my question is really whether in its current form the reasoning is valid.
• The fact that (k-k) / h is 0/0 when h=0 is substituted does not mean that lim h>0 (k-k) / h has a final value of 0/0. Note that the division property of limits does not apply if the limit of the denominator function is zero, so lim h>0 (k-k) / h should not be thought of as [lim h>0 (k-k)] / (lim h>0 h), which would be 0/0.
The fact that that (k-k) / h is 0/0 when h=0 is substituted does mean that more work (such as algebraic simplification or L'Hopital's Rule) is required to find lim h>0 (k-k) / h. Note that as long as h is not exactly zero, (k-k) / h simplifies to 0/h, which then simplifies to 0. So lim h>0 (k-k) / h = lim h>0 0 = 0. The value of this limit is well defined and equals 0.
• What does the "h" in the formula mean?
• `h` is the result of `(x+h)-x`, where `(x+h)` is a point/value infinitesimally close to `x`.
• I know how to calculate the first derivative,but I want to know is there a way to calculate " first half derivative"?
(1 vote)
• There is such a thing, and the study of non-integer derivatives is called fractional calculus. The wikipedia page for Fractional Calculus gives a bit of a summary of it, but it requires a lot more abstraction and mathematical background than standard calculus.
• Do you divide x/y or y/x to find your constant multiple?Do you put the y number inside the dividing house or put x number inside?
• A "constant multiple" is just a number. You do not really "find" it, you define it.

Like the graph of y = 3... what is that graph? 3 is a CONSTANT, it's just "a number", when you say a constant it's the opposite of a variable, a variable can change, a constant is always that number. 3 is always 3. Can you change your constant? Sure, you could say y = 4, y = 8, etc.

Anyway, the graph of y = 3 is just a horizontal line. So when we differentiate it the result is 0. Remember we associated a differentiate function as the "slope" of that function at a certain point so far. Well... what is the slope for a horizontal line? Nothing... it's 0, it has no slope, no rise/run, nada. So when we differentiate a constant we find that value is 0. Hence: f(x) = k will be f'(x) = 0 when differentiate, assuming k is an arbitrary constant (meaning it can be any constant you want, 2, 5, 84385, whatever, they all have the same slope of 0).

You can apply the difference quotient, but you'll find it's rather lackluster as Sal did:

f(x+h) - f(x)
-----------------
h

Remember, f(x) is just a number, so let's replace it with any number we want, say 3:

3 - 3
-------
h

Which is 0/h, and hence the difference quotient is 0, the slope is 0. A differentiated constant is just 0.
• Okay, so as always thanks beforehand for answering... But I don't get this...why is taking the derivative of a constant equal to 0 (well I get why because 0 has no slope) but why then does a derivative of a variable like x equal to 1? Isn't X a number as well? so...it should be treated just like a constant would yes? What makes a variable like X different than just a number like pi, or 4 or 937?
(1 vote)
• What is the definition of differentiation?
What is the definition of intefration?
(1 vote)
• what is the l hospital rule?
(1 vote)
• Isn't a perpendicular line also passes through one point ?
• can someone help me differentiate y=5/1-3x
• First, rewrite as y = 5 (1-3x)^-1, then apply the power and solve. You should get y' = 15(1-3x)^-2 or y' = 15/(1-3x)^2
(1 vote)

## Video transcript

- [Voiceover] So these are both ways that you will see limit-based definitions of derivatives. This is usually if you're thinking about the derivative at a point, here if you're thinking about the derivative in general, but these are both equivalent. They're both based on the slope of a tangent line, or the instantaneous rate of change, and using these, I wanna establish some of the core properties of derivatives for us. And the first one that I'm going to do will seem like common sense, or maybe it will once we talk about it a little bit, so if f of x, if our function is equal to a constant value, well then, f prime of x is going to be equal to zero. Now why does that make intuitive sense? Well, we could graph it, we could graph it, so if that's my y-axis, that's my x-axis. If I wanted to graph y equals f of x, it's gonna look like that, where this is at the value y is equal to k, so this is y is equal to f of x. Notice no matter what you change x, y does not change. The slope of the tangent line here, well frankly, is the same line, it has a slope of zero. No matter how, y is just not changing here, and we could use either of these definitions to establish that even further, establish it using these limit definitions, so let's see, the limit as h approaches zero of f of x plus h, well no matter what we input into our function, we get k, so f of x plus h would be k minus f of x. Well, no matter what we put into that function, we get k over h, well, this is just going to be zero over h so this limit is just going to be equal to zero. So f prime of x for any, for any x, the derivative is zero, and you see that here, that the slope of the tangent line for any x is equal to zero. So if someone walks up to you on the street and says, "Okay, h of x, h of x, h of x, is equal to pi, "what is h prime of x?" Well, you say, well, pi, that's just a constant value that the value of our function is not changing as we change our x, the slope of the tangent line there, the instantaneous rate of change, it is going to be equal to zero.