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Proof of the Product Rule

  1. Mar 16, 2012 #1
    Hello there!

    I understand the product rule and how/why it works, and the proof makes sense,
    but just out of curiosity, why would it be incorrect on a mathematical/application level to say that d/dx f(x)g(x) = f'(x)g'(x)

    I know that it's wrong, and the product rule is the one that we're supposed to use, but I don't understand on a practical level WHY it's wrong.

    Please help! Thank you :)
     
  2. jcsd
  3. Mar 16, 2012 #2

    Mark44

    Staff: Mentor

    Because it doesn't give the right answer!

    Here's an example: let f(x) = x2, and g(x) = x2. Then by your version of the product rule, d/dx(f(x) g(x)) = 2x * 3x2 = 6x3.

    OTOH, f(x)*g(x) = x5, so d/dx(f(x)*g(x) ) = 5x4.

    The real product rule gives this result, which you can confirm.
     
  4. Mar 16, 2012 #3
    Well, product rule is something which can be derived in few steps.

    Let y= f(x) and z= g(x)
    Let h(x)=f(x)g(x)
    For any x, h(x)=f(x)g(x)=yz
    To map changes, we say that if x increases by dx, let change in y be dy and change in z be dz.
    So h(x+dx)= f(x+dx)g(x+dx)=(y+dy)(z+dz)

    =yz+ydz+zdy+dz*dy
    =yz+ydz+zdy as last term is v v v small .

    Now h'(x) is defined as
    ((h+dx)-h(x))/dx

    What do you see?
     
  5. Mar 16, 2012 #4
    Haha, that was simpler than I thought! :D Thank you so much!
     
  6. Mar 16, 2012 #5
    Oh there's a cool visualization.

    Take a rectangle with width w and height h. Draw a picture of it.

    Now increase the width by delta-w and the height by delta-h. Draw the picture and you'll see you now have now increased the area wh by the sum of three new rectangles. As delta-w and delta-h get close to zero, the corner rectangle with area (delta-h)(delta-w) gets MUCH smaller than the other two rectangles so you can ignore it ... and voila, you have the product rule.

    I googled around and found a picture ...

    proof-2-of-product-rule.JPG
     
  7. Mar 16, 2012 #6
    Your welcome. :-)

    Btw the last post is amazing.
    Nice one Steve
     
  8. Mar 16, 2012 #7
    well, I guess it depends on how you want to interpret(define) f'(x). If you interpret f'(x) as the rate of change of f with respect to x at a given point which is associated with the slope of the tangent line passing through the graph of f at that point (That's how the derivative of a function is defined as you know), then you could simply check that (fg)'=f'g' does not tell you the rate of change of (fg)(x) with respect to x. If you wanna see this, just take two arbitrary functions, like f(x)=x and g(x)=1+x^2, then (fg)(x)=x(1+x^2). Now graph (fg)(x) and try to manually calculate the slop of fg at the point x=0 for instance. That means choose a small interval centered at x=0 and calculate the line that passes through the beginning and ending points of the interval on the graph. If you do this, you'll see that the slope of the tangent line at x=0 would be 1, but if you defined (fg)'=f'g' then you would get 0 which is not certainly not true.(It's also very easy to check if you look at the graph of fg: http://www.wolframalpha.com/input/?i=x%281%2Bx^2%29)
    If you still doubt that it's very reasonable to define (fg)'=(f').g+(g').f then you could directly obtain the formula using the definition of derivative.
     
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