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Math Help - Quick Quotient Rule Question

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    Quick Quotient Rule Question

    Hello all, I have been having some trouble grasping the quotient rule. Here's a sample problem, and I'll point out where I get tripped up.

    f(x)=\frac{12.6(4.8^{x})}{x^2}

    1) f'(x)=\frac{d}{dx}[{12.6(4.8^{x})}]*{x^2}-{12.6(4.8^{x})}*\frac{d}{dx}({x^2}) all over {(x^2)^2} (I couldn't find the right LaTeX command to divide the entire function by {(x^2)^2})

    2) f'(x)={12.6*(4.8^{x}ln4.8)}]*{x^2}-{12.6(4.8^{x})}*({2x}) all over {(x^4)}

    3) I divided {(x^4)} from {x^2} in the numerator, and then combined like terms. f'(x)={12.6{x^{-2}}*(4.8^{x}ln4.8)}]}-{25.2x(4.8^{x})}

    ----I think I see where I went wrong. When I divide {(x^4)} in the denominator, I need to also do it to \frac{2x}{(x^4)}?; the book's answer is f'(x)={12.6(4.8^{x}ln4.8)}}{x^{-2}}-{25.2(4.8^{x})}{x^{-3}}.

    I guess what I'm asking is if my answer in part 3 is correct assuming the book's is just another way to simplify the same problem?
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    Re: Quick Quotient Rule Question

    Quote Originally Posted by AZach View Post
    I think I see where I went wrong. When I divide {(x^4)} in the denominator, I need to also do it to \frac{2x}{(x^4)}?; the book's answer is f'(x)={12.6(4.8^{x}ln4.8)}}{x^{-2}}-{25.2(4.8^{x})}{x^{-3}}.

    I guess what I'm asking is if my answer in part 3 is correct assuming the book's is just another way to simplify the same problem?
    Your answer in part 3 is incorrect, for the reason that you stated. In general, you can cancel factors of the numerator and denominator. But if you want to cancel a factor from an individual term, that is not allowed. You must cancel from every term in the numerator.

    Continuing from step 2, we have

    f'(x) = \frac{12.6x^2\left(4.8^x\ln4.8\right) - 25.2x\left(4.8^x\right)}{x^4}

    f'(x) = \frac{x^4\left[12.6x^{-2}\left(4.8^x\ln4.8\right) - 25.2x^{-3}\left(4.8^x\right)\right]}{x^4}

    f'(x) = 12.6x^{-2}\left(4.8^x\ln4.8\right) - 25.2x^{-3}\left(4.8^x\right)
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