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Proving the product rule

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.B (LO)
,
FUN‑3.B.1 (EK)
Proving the product rule for derivatives.
The product rule tells us how to find the derivative of the product of two functions:
ddx[f(x)g(x)]=ddx[f(x)]g(x)+f(x)ddx[g(x)]=f(x)g(x)+f(x)g(x)\begin{aligned} \dfrac{d}{dx}[f(x)\cdot g(x)]&=\dfrac{d}{dx}[f(x)]\cdot g(x)+f(x)\cdot\dfrac{d}{dx}[g(x)] \\\\ &=f'(x)g(x)+f(x)g'(x) \end{aligned}
The AP Calculus course doesn't require knowing the proof of this rule, 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.

Without further ado, we present to you the proof!

Khan Academy video wrapper
Product rule proofSee video transcript

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  • leaf blue style avatar for user Russell Robert Anthony Alson
    Is there reasons why there are many proofs?
    (16 votes)
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  • male robot johnny style avatar for user siddarth011089
    According to the Limit rules lim x->a f(x).g(x) = lim x->a f(x) . lim x->a g(x) . But why when the derivative is calculated this rule is violated?
    (9 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      Interesting question! The derivative of a function is not just the limit of the function, but rather is a limit of something more complicated that is related to the function.

      Note that the derivative of f(x)g(x) is limit h-->0 of [f(x+h)g(x+h) - f(x)g(x)]/h,
      the derivative of f(x) is limit h-->0 of [f(x+h) - f(x)]/h,
      and the derivative of g(x) is limit h-->0 of [g(x+h) - g(x)]/h.

      In general, [f(x+h)g(x+h) - f(x)g(x)]/h is not the product of [f(x+h) - f(x)]/h and [g(x+h) - g(x)]/h, so we can't just use the product property of limits to conclude that the derivative of f(x)g(x) is the product of the derivatives of f(x) and g(x).

      Have a blessed, wonderful day!
      (19 votes)
  • blobby green style avatar for user Harikesh
    At I did not understand why d/dx[f(x)*g(x)] =

    lim [f(x+h)g(x+h) - f(x)g(x)]/h
    h->0
    (6 votes)
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  • aqualine seed style avatar for user Fibonacci
    Why does lim h->0 f(x + h) is equal to f(x) ?
    (3 votes)
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    • leaf green style avatar for user kubleeka
      Because if a function is continuous, then the limit of the function is equal to the function of the limit. That is,
      lim_{x-> c} f(x)=f(lim_{x->c} x)=f(c)

      In this case, we have assumed f to be differentiable, and therefore continuous, so lim_{h->0} f(x+h)=f(x+0)=f(x).
      (5 votes)
  • blobby green style avatar for user Dan
    How does Sal go from step 2 to step 3 (minute to )? I mean, how is it concluded that lim h->0 [f(x+h)*(g(x+h)-g(x))/h] = lim h->0 [f(x+h)]*lim h->0 [(g(x+h)-g(x))/h]?

    In other words, if the limit of two functions multiplying each other is lim h->0 [f(x)g(x)] = f’(x)g(x) + f(x)g’(x), then why does Sal says that “the limit if the products is the product of the limits”? Wouldn’t we fall into a redundance in the proof? Please help!
    (2 votes)
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    • leaf green style avatar for user kubleeka
      You're confusing the product rule for derivatives with the product rule for limits. The limit as h->0 of f(x)g(x) is
      [lim f(x)][lim g(x)], provided all three limits exist. f and g don't even need to have derivatives for this to be true.

      The derivative of f(x)g(x) if f'(x)g(x)+f(x)g'(x). This is not the same as the limit.
      (5 votes)
  • blobby green style avatar for user gunank312
    at , did we assume the two functions be on a same graph or something? i did not get the step... why do we multiply (f.g)(x+h) and subract it by f.g(x) and have only one h in the denominator? how to visualise two functions of distinct graphs, f(x) and g(x) being multiplied, that too of their slopes?

    thanks
    (2 votes)
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    • leaf grey style avatar for user Kshitij
      Let the function H(x) = f(x)*g(x). H(x) can be plotted on the same graph as f(x) and g(x) because all 3 of them are functions of "x" and only "x".

      We have to find the derivative of H(x), which is
      lim[h-->0] (H(x+h) - H(x))/h

      H(x+h) is f(x+h)g(x+h) and H(x) is f(x)g(x), the limit can be written as
      lim[h-->0] (f(x+h)*g(x+h) - f(x)*g(x))/h
      (2 votes)
  • piceratops ultimate style avatar for user Aryan.patel
    According to the Limit rules lim x->a f(x).g(x) = lim x->a f(x) . lim x->a g(x) . But why when the derivative is calculated this rule is violated?
    (2 votes)
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  • starky ultimate style avatar for user Ved Sharma
    When the f(x+h) and g(x) were factored out of the numerator, why were there no parentheses? The way it was written, it seems like the f(x+h) and g(x) would just be multiplied to the numerator.
    (1 vote)
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  • leaf green style avatar for user Sahana K.
    At , why wouldn't the denominator be h^2? If you multiply the derivative formula for f(x) and the derivative formula for g(x), wouldn't the h become h^2?
    (2 votes)
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  • blobby green style avatar for user Sohta Inoguchi
    At and , I understand that "lim as h approaches 0, (g(x+h)-g(x))/h" and "lim as h approaches 0, (f(x+h)-f(x))/h" are g'(x) and f'(x) respectively, but if zero is plugged into h, aren't they "undefined" since their denominators are zero?
    (1 vote)
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