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## Differential Calculus

### Course: Differential Calculus > Unit 2

Lesson 7: Derivative rules: constant, sum, difference, and constant multiple# Justifying the basic derivative rules

AP.CALC:

FUN‑3 (EU)

, FUN‑3.A (LO)

, FUN‑3.A.2 (EK)

The basic derivative rules tell us how to find the derivatives of constant functions, functions multiplied by constants, and of sums/differences of functions.

Constant rule | start fraction, d, divided by, d, x, end fraction, k, equals, 0 | |

Constant multiple rule | start fraction, d, divided by, d, x, end fraction, open bracket, k, dot, f, left parenthesis, x, right parenthesis, close bracket, equals, k, dot, start fraction, d, divided by, d, x, end fraction, f, left parenthesis, x, right parenthesis | |

Sum rule | start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, plus, g, left parenthesis, x, right parenthesis, close bracket, equals, start fraction, d, divided by, d, x, end fraction, f, left parenthesis, x, right parenthesis, plus, start fraction, d, divided by, d, x, end fraction, g, left parenthesis, x, right parenthesis | |

Difference rule | start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, minus, g, left parenthesis, x, right parenthesis, close bracket, equals, start fraction, d, divided by, d, x, end fraction, f, left parenthesis, x, right parenthesis, minus, start fraction, d, divided by, d, x, end fraction, g, left parenthesis, x, right parenthesis |

The AP Calculus course doesn't require knowing the proofs of these rules, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

## Let's first see why the constant rule is true.

## Now let's prove the constant multiple and sum/difference rules.

## Want to join the conversation?

- Is there a graphical representation for these proofs ? This is because I dont know the limit function ( but I will study about it later) .(11 votes)
- you can watch 3blue1brown's video series called "essence of calculus" on you tube.(43 votes)

- From the first video, isn't lim h -> 0 0/h indeterminate (0/0), not equal to 0? You get the same thing when you use the alternate form of the definition of a derivative as well.(15 votes)
- You get an indeterminate form if you just substitute 0 for h, but that doesn't mean the limit doesn't exist. It just means the numerator and denominator both go to 0.

As x goes to 0, x/x doesn't go to 0/0, it's constant at 1.(5 votes)

- Is:

equivalent to`d/dx * [f(x)+g(x)]`

`d/dx * (f(x)+g(x))`

... and if so, what is the deal with "[]" instead of parantheses?(4 votes)- They're the same. We just use the brackets to avoid confusion with the parentheses on f(x) and g(x).(19 votes)

- Hi, could someone please give me a real world example of a function that is the sum of 2 other functions? And why you would want to find its derivative? I am enjoying these videos and think Sal does a good job describing all the steps, but just like when I was at High School, I am having touble appreciating the importance of these concepts as I don't understand how they are applied in the real world. Thanks.(5 votes)
- Lets say that you have two water faucets that give water at two different rates. The first faucet gives water at 1 liter per minute and the other gives water at 2 liter per minute. Lets say you want to find how much water in total is released after 5 minutes. You would create two functions as follows:

Faucet 1: F1(m) = 1m (m being minutes and F1 the number of liters of water).

Faucet 2: F2(m) = 2m. (F2 is the number of liters).

So to find the liters of water after 5 minutes, you would plug 5 in for each function and then sum the answers. This is the same as creating a function like (F1 + F2)(m). Notice that this is the same as F1(m) + F2(m). Hope this helps!(13 votes)

- In the first video, we have the limit as x approaches 0 of (k - k) / h. The k - k evaluates to 0, and this leads Sal to write that the entire limit is equal to 0.

Is it always the case that when the numerator is 0, the limit is 0, regardless what the denominator is?(4 votes)- Not necessarily. When the numerator is a constant 0 (such as in the case of
`(k-k)/h`

, the limit evaluates to 0. However, if the numerator involves a variable and just happens to evaluate to 0, you also need to consider the denominator.

For example, when evaluating the limit as x approaches 0 of`x/sin(x)`

, the numerator evaluates to 0. However, if you graph it out, you can see that the limit is actually 1 instead of 0, because the denominator also evaluates to 0, and 0/0 is indeterminate. In this case, we have to consider other methods, such as l'Hopital's rule instead of directly evaluating.(8 votes)

- in the first video, isn't:

lim h->0 (k-k/h)

equal to

(lim h->0 (k-k))/(lim h->0 (h))

which is equal to

0/0

which is undefined?

Also how do you switch to full player? I'm having trouble running the video.(4 votes)- lim (a/b)=(lim a)/(lim b) only when all three limits are defined and lim b is not 0. If lim b is 0, lim (a/b) may or may not be defined. We generally have to do more work to find out.(5 votes)

- In the second video, the proof for the constant multiple rule for derivatives more or less shifts the burden of proof onto the corollary rule for limits (the constant multiple rule for limits). These two rules seem to be basically different versions of the same thing, one for when dealing with derivatives, one for when dealing simply with limits. Could Sal provide us with the underlying proof for the constant multiple rule for limits?(4 votes)
- Suppose the limit as x→c of f(x)=L. We want to show that the limit as x→c of k·f(x)=kL.

Recall the definition of a limit: for all ε>0, there exists δ>0 such that for all x, |x-c|<δ implies |f(x)-L|<ε. We've assumed that this is true for this c, f, and L, for any ε. Now we'll show it for kf(x) and kL, using an arbitrary ε' and δ'.

So choose an ε' such that 0<ε'<ε/k, and let δ'=δ. Then if |x-c|<δ, we have

|k·f(x)-kL|

=k·|f(x)-L|<k·ε'<ε.

So the limit as x→c of k·f(x)=kL, which was what we wanted.(5 votes)

- I want to know what's the difference between d/dx & dy/dx & f'(x)(2 votes)
- d/dx is a verb: take the derivative of this function

dy/dx is a noun: this is the derivative of this function y with respect to x

f'(x) is the derivative of the function f(x)(6 votes)

- Even after reading the comments, I'm still a bit confused about why lim h-->0 (k-k/h) equals 0. Wouldn't it be equal to 0/0 and therefore indeterminate? I understand the constant rule looking at it graphically but the algebraic part of the proof confuses me. I'd be very grateful to anyone who could clear this up.(3 votes)
- One way to do it is by using L'Hopital's rule, but since it isn't taught yet you can rely on using the graph of the function f(h) = 0/h as h approaches zero.

When the limit approaches from the left all the values are zero as 0/(anything non zero) is zero and similarly for the right-hand side the value is always zero hence the limit is zero.(4 votes)

- The derivative is the slope of the tangent line to a point on a function, right? So wouldn't the tangent line to f(x) = k be a vertical line, which has no slope?

On the other hand, algebraically it makes sense that f'(x) = 0, so I'm very confused.(3 votes)- The tangent line at any point along a horizontal line is another horizontal line, not a vertical line. The slope of a horizontal line is 0, so the derivative at any point along a horizontal line will be 0.(3 votes)