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Find g'(x)
I thought that you use the product rule but I have the answer and I am not getting the same thing.
The answer is
g'(x)=3/2[4x^2+37x-77]^1/2 * (8x+37)

I keep getting just 8x+37 when I do the product rule. What am I missing?

  • Calculus -

    If the question is the way you stated it, then the derivative is indeed 8x - 37

    judging from the answer that "they" give your question must have been

    g(x)=[(4x-7)(x+11)]^(3/2), (they expanded the inside first).

  • Calculus -

    You have to use the FOIL rule that stands for first,outer, inner and last.
    So look at the two equatons:
    (4x-7)(x+11) Firt stands for the multiplication of 4x and x in which are the first number of the equation,next outer means that you now multiply the outer numbers being 4x and 11. Thirdly, inner stands for the multiplication of the two inner numbers being -7 and x. At last you multiply the last two numbers of each equation being -7 an 11. If you do it in order you will get 4x^2+44x-7x-77 than you simplify the two like terms being 44x-7x in which it gives you 37x, So your answer will be 4x^2+37x-77.

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