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Thread: Supposedly easy derivative:

  1. #1
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    Supposedly easy derivative:

    Hello all,

    I'm in charge of finding the derivative of $\displaystyle 1 \over 4x-3$.

    I first used the quotient rule, which came out to be

    $\displaystyle 0 * (4x-3) - ((4x-3) * 1) \over (4x-3)^2$

    and simplified to

    $\displaystyle - 4x-3 \over (4x-3)^2$


    However, this isn't complete. I understand that I need to use the chain rule on the numerator next, according to Wolfram-Alpha, but I don't understand why I need the chain rule since there's only 1 function. Can anyone shed some light on this?

    Thank you in advance!
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  2. #2
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    Use the law of exponents that says $\displaystyle \dfrac{1}{a} = a^{-1}$ to rewrite $\displaystyle \dfrac{1}{4x-3} = (4x-3)^{-1}$

    You can the use the chain rule
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  3. #3
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    Oh wow! Haha.

    Now that you show that, it's pretty easy to see. I'd imagine I should do this for all fractions with 1 in the numerator correct?

    It makes complete sense:

    $\displaystyle (4x-3)^{-1}$

    $\displaystyle (4x-3)^{-1} = (4x-3)^{-2} * 4$

    $\displaystyle 4 \over (4x-3)^2$


    EDIT: I should say, a constant in the numerator
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  4. #4
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    e^(i*pi)'s Avatar
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    Yes, but bear in mind that it will not be affected by the exponent. That is $\displaystyle \dfrac{2}{x-1} = 2(x-1)^{-1} \neq (2(x-1))^{-1}$


    Edit. Your answer is missing a minus sign out front.
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  5. #5
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    Quote Originally Posted by sinx View Post
    Hello all,

    I'm in charge of finding the derivative of $\displaystyle 1 \over 4x-3$.

    I first used the quotient rule, which came out to be

    $\displaystyle 0 * (4x-3) - ((4x-3) * 1) \over (4x-3)^2$
    No, that's not correct. The quotient rule says that
    $\displaystyle \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$
    Here f(x)= 1 so f'(x)= 0, g(x)= 4x- 3 so that g'= 4.
    $\displaystyle \frac{0(4x-3)- (4)(1)}{(4x- 3)^2}$

    and simplified to

    $\displaystyle - 4x-3 \over (4x-3)^2$


    However, this isn't complete. I understand that I need to use the chain rule on the numerator next, according to Wolfram-Alpha, but I don't understand why I need the chain rule since there's only 1 function. Can anyone shed some light on this?

    Thank you in advance!
    No, not "use the chain rule on the denominator". But you did forget to differentiate 4x- 3.
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