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Homework Help: How in the blazes is this the derivative of f(x)?

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    f(x)=(2x+1)^11 (5x-1)^9

    2. Relevant equations

    3. The attempt at a solution
    f'(x)=[(2x+1)^10 (5x-1)^8] * (200x+23)
    Where does the +23 come from, I don't get it. I get the rest.
    Last edited: Oct 25, 2009
  2. jcsd
  3. Oct 25, 2009 #2


    Staff: Mentor

    Your function and it's derivative are difficult to comprehend, as you have an error in your LaTeX.
    I believe this is what you meant:
    [tex]f(x)=(2x+1)^{11} (5x-1)^9[/tex]
    [tex]f'(x)=[(2x+1)^{10} (5x-1)^8] * (200x+23)[/tex]

    If you have an exponent that consists of more than one character, you have to put braces, {}, around it.

    Using the product rule, f'(x) = 11*2*(2x + 1)10*(5x - 1)9 + (2x + 1)11*9*5*(5x - 1)8
    = 22*(2x + 1)10*(5x - 1)9 + 45*(2x + 1)11*(5x - 1)8
    = (2x + 1)10*(5x - 1)8[22(5x - 1) + 45(2x + 1)]
    = (2x + 1)10*(5x - 1)8[110x - 22 + 90x + 45]
    = (2x + 1)10*(5x - 1)8[200x + 23]
  4. Oct 25, 2009 #3
    EDIT: Way too late. Disregard this, Mark44 beat me to it.

    You use the product rule to get:
    [tex]f'(x) = (5x-1)^9\frac{d}{dx}(2x+1)^{11} + (2x+1)^{11} \frac{d}{dx}(5x-1)^9[/tex]
    Use the chain-rule to calculate these two derivatives. Then with a little bit of rearranging you should get the desired result. As to where the 23 comes from it occurs naturally when you do the calculation, or a bit more concretely:
    [tex](5x-1)11(2) + (2x+1)9(5) = 110x+90x+45-22 = 200x+23[/tex]
    which you shall find when you do the calculations.

    I don't see how you can get the rest of the answer unless you made a mistake. Try showing how you got the rest, but not the +23.
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