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Math Help - Using quotient and chain rules for derivatives

  1. #1
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    Using quotient and chain rules for derivatives

    Hey so I have a calculus assignment and im not wrapping my head around many of the concepts for derivatives.

    My quotient rule question:

    a) Use the definition of the derivative to find f'(x) for f(x) = (1/3x-4).
    b) Use the result from a) to fi nd the equation of the tangent line to f(x) at the point (2, 1/2 ).

    For this I believe I have answered the first part, with

    f(x)=(1/3x-4)

    f'(x)=(f(x)*g'(x)-f'(x)g(x))/f(x)^2

    =-3/(3x^2-24x+16)

    I'm don't know what to do to continue to part be.
    Thanks very much for any help you might be able to offer.
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  2. #2
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    Quote Originally Posted by commerce-calc View Post

    = \frac{-3}{3x^2-24x+16}

    I'm don't know what to do to continue to part be.
    Thanks very much for any help you might be able to offer.
    Substitute the Point 2 into \frac{-3}{3x^2-24x+16} , which will give you the gradient of the tangent line.

    Now, for y-coordinate, subtitute the point x=2 into f(x) i.e. \frac{1}{3x-4}, which will give you the y-coordinate.

    Then simply put the data into the either 2 formulas Point-Slope or Slope-Intercept, I recommend point-slope.

    and there you go, hope you understood
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  3. #3
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    awesome. that actually makes sense im not used to that haha.

    so i came up with the equation for the tangent line:

    y=3/10x-01

    sound maybe right? thanks for clearing that up for me
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  4. #4
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    Happy to help, just going along these things
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  5. #5
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    Quote Originally Posted by commerce-calc View Post

    =-3/(3x^2-24x+16).
    Just noticed this, 3x^2 shouldn't be 3x, it should be 9x^2, correct it, and make the equation again, this time you'll get the right one, sorry for not pointing out earlier
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  6. #6
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    haha no problem. and thank you very much for noticing the error, fixed and good. thanks again for the help.
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