Is there a way.....?

is there a way to prove Leibniz's law (the product rule of calculus) without using derivations?

7 Answers

  • 1 decade ago
    Favorite Answer

    Don't know......haven't studied this topic till yet....!

  • Ted
    Lv 4
    1 decade ago

    I don't know exactly what you mean. A proof is itself a derivation. See the link below. They explain Liebniz insight. This isn't a proof though; it's just an insight. He said that the differential:


    was equal to:

    (u+du)(v+dv) - u*v

    That is, it's:

    u*dv + v*du + du*dv

    The u*v was subtracted because it doesn't contribute to the differential. Liebniz then said that the du*dv was neglible, so he dropped it which gave:

    d(uv) = u*dv + v*du

    which is the differential form of the product rule. "Divide" both sides by "dx" to get the more familiar form:

    d(uv)/dx = u*dv/dx + v*du/dx

    However, this really isn't a "proof." The product rule can be proved pretty simply by using the definition of a derivative. The deriviative of f(x)*g(x) at a point x1 is defined as the limit of:

    ( f(x2)*g(x2) - f(x1)*g(x1) )/ (x2-x1)

    as x2->x1. However, it's easy to see that you can rewrite this ratio as (i.e., change (a-b)/c to a/c - b/c):

    f(x2)*g(x2)/(x2-x1) - f(x1)*g(x1)/(x2-x1)

    Now add -g(x2)*f(x1)/(x2-x1) + g(x2)*f(x1)/(x2-x1) to the expression. This is equivalent to adding 0. You're just adding and subtracting the same thing, like adding d and then subtracting d. You get:

    g(x2)*f(x2)/(x2-x1) - g(x2)*f(x1)/(x2-x1) - f(x1)*g(x1)/(x2-x1) + f(x1)*g(x2)/(x2-x1)

    but you can factor out the g(x2) from the left two terms and the f(x1) from the right two terms and get:

    g(x2)*( f(x2)-f(x1) )/(x2-x1) + f(x1)*( g(x2)-g(x1) )/(x2-x1)

    Now, remember we're taking the limit of all of this. Because we assumed that g was differentiable, we know that g(x2)->g(x1) as x2->x1. So we replace g(x2) with g(x1). The two ratios are just the definition of the derivatives f'(x) at x1 and g'(x) at x1. So substituting in g(x1), f'(x1), and g'(x1), we get:

    g(x1)*f'(x1) + f(x1)*g'(x1)

    That's what we want. That's a rigorous way to prove the product rule, and I don't think it's that much more difficult. It's probably better to just use the definition of the derivative.

  • 1 decade ago

    wel yes if u r asking for a proof of dy/dx=(dy/du)*(du/dx).

    u would start with elemntals (simple delta stuff) of x and y n their ratio. just insert two du elements to get the above expression in terms of delta things and then just let delta x go to zero, then one would become dy/du and other du/dx as du also goes to zero as dx does.if this does not saisfy u or u want a clarification just IM or mail me.regards

  • 1 decade ago

    For simplicity lim will mean limit as h -->0

    Definition of derivative:

    f ' = lim ( f(x + h) – f(x) )/h

    let u(x) = f(x)g(x)

    u' = lim ( u(x + h) – u(x) )/h

    = lim ( f(x + h)g(x + h) – f(x)g(x) )/h

    Add and subtract f(x)g(x + h)

    = lim ( f(x + h)g(x + h) – f(x)g(x) + f(x)g(x + h) - f(x)g(x + h) )/h


    = lim ( f(x + h)g(x + h) - f(x)g(x + h) + f(x)g(x + h) – f(x)g(x) )/h

    = lim ( [ f(x + h) - f(x) ]g(x + h) + f(x)[ g(x + h) – g(x) ] )/h

    Distribute the limit

    = lim ( [ f(x + h) - f(x) ] )/h lim g(x + h)/h + f(x)lim[ g(x + h) – g(x) ] /h

    The term lim g(x + h)/h = g(x)

    So we have

    = g(x)lim ( f(x + h) - f(x) )/h + f(x)lim (g(x + h) – g(x) ) /h

    Which is f 'g + fg'

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  • Anonymous
    1 decade ago

    No..I don't believe so. Sorry!

  • 1 decade ago

    you bet

  • 1 decade ago

    dont think so

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