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Math Help - Finding the derivative of a large function. Help with the order of operations.

  1. #1
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    Finding the derivative of a large function. Help with the order of operations.

    Differentiate
    (9x -3x^2)^3 cos(4x^3 3x)
    e^(7-12x) ln(-x^2 + 3)
    With respect to x

    I don't want the solution, what i'm after is some help in how to tackle this differentiation question.

    This is what i plan to do:
    1. Find dy/dx for (9x-3x^2)^3 [make this part A]
    2. find dy/dx for cos(4x^3 - 3x) [make this part B]
    3. find dy/dx for e^(7 - 12x) [make this part C]
    4. find dy/dx for ln(-x^2 + 3) [make this part D]
    5. product rule A and B [make this part E]
    6. product rule C and D [make this part F]
    7. quotient rule E and F


    Does this look right?
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  2. #2
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    \frac{d}{dx} \left(\frac{uv}{wz}\right) = \frac{(wz)(uv'+vu')-(uv)(wz'+zw')}{(wz)^2}
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  3. #3
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    Thank you so much. I have never seen this version of the quotient rule before.
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  4. #4
    A Plied Mathematician
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    As an alternative, you could use logarithmic differentiation.
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  5. #5
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    Quote Originally Posted by ezzab69 View Post
    Thank you so much. I have never seen this version of the quotient rule before.
    It is simply \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{fg} with f= uv and g= wz. Of course, f'= (uv)'= u'v+ uv' and g'= (wz)'= w'z+ wz'
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  6. #6
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    Re: Finding the derivative of a large function. Help with the order of operations.

    Quote Originally Posted by HallsofIvy View Post
    It is simply \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{fg} with f= uv and g= wz. Of course, f'= (uv)'= u'v+ uv' and g'= (wz)'= w'z+ wz'
    Just quickly clarify, you're using the quotient rule right? So shouldn't it be: \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2} ?
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  7. #7
    Member sbhatnagar's Avatar
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    Re: Finding the derivative of a large function. Help with the order of operations.

    Quote Originally Posted by ezzab69 View Post
    Just quickly clarify, you're using the quotient rule right? So shouldn't it be: \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2} ?
    You are correct.
    \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}
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