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Justifying the basic derivative rules

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 ruleddxk=0
Constant multiple ruleddx[kf(x)]=kddxf(x)
Sum ruleddx[f(x)+g(x)]=ddxf(x)+ddxg(x)
Difference ruleddx[f(x)g(x)]=ddxf(x)ddxg(x)
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.

Khan Academy video wrapper
Proof of the constant derivative ruleSee video transcript

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

Khan Academy video wrapper
Proofs of the constant multiple and sum/difference derivative rulesSee video transcript

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  • mr pink red style avatar for user Nikhil Raju
    Is there a graphical representation for these proofs ? This is because I dont know the limit function ( but I will study about it later) .
    (12 votes)
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  • purple pi purple style avatar for user Lavie
    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.
    (16 votes)
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    • leaf green style avatar for user kubleeka
      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)
  • blobby green style avatar for user tamnor94
    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.
    (6 votes)
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    • hopper cool style avatar for user obiwan kenobi
      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!
      (19 votes)
  • aqualine ultimate style avatar for user sebastian nielsen
    Is:
              d/dx * [f(x)+g(x)]
    equivalent to
              d/dx * (f(x)+g(x))

    ... and if so, what is the deal with "[]" instead of parantheses?
    (3 votes)
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  • leaf green style avatar for user S. ABBASI
    I want to know what's the difference between d/dx & dy/dx & f'(x)
    (4 votes)
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  • mr pink green style avatar for user chris
    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)
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    • hopper jumping style avatar for user Tyler Tian
      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)
  • leafers ultimate style avatar for user Andrew Escobedo
    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)
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  • leafers tree style avatar for user Robert
    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)
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    • leaf green style avatar for user kubleeka
      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)
  • starky sapling style avatar for user kesteele1221
    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.
    (4 votes)
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    • leaf grey style avatar for user Kshitij
      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)
  • blobby green style avatar for user holasoymas54
    is d f(x)/dx , dy/dx, f'(x) same thing ?
    (3 votes)
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    • leaf green style avatar for user Tanner P
      Yes, they all mean the derivative.

      The dy/dx notation is useful because it emphasizes the idea that the derivative is a small change in the output divided by a small change in the input.

      And f'(x) is useful because it shows that the derivative itself is a function.
      (3 votes)