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Math Help - Chain Rule - Book Example

  1. #1
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    Post Chain Rule - Book Example

    I am having trouble understanding a portion of the textbook example. It's seems to be mostly with the simplifing of the answer.

    Find dy/dx if y = u^3 - 3u^2 + 1 and u = x^2 + 2

    dy/du = 3u^2 - 6u and du/dx = 2x

    it follows that dx/dy = (3u^2 - 6u)(2x). *I'm fine so far, keep going*

    We are substituting x^2 + 2 for u in the expression for dy/dx giving
    [3(x^2+2)^2 - 6(x^2+2)](2x)

    *I understand everything above but the next step confuses me*

    6x(x^2+2)[(x^2+2)-2]
    I don't know what steps they took to arrive at that equation. Could you possible show me the 'longer' process?

    It goes on to give:
    6x(x^2+2)(x^2)=
    6x^3(x^2+2)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by becky
    I am having trouble understanding a portion of the textbook example. It's seems to be mostly with the simplifing of the answer.

    Find dy/dx if y = u^3 - 3u^2 + 1 and u = x^2 + 2

    dy/du = 3u^2 - 6u and du/dx = 2x

    it follows that dx/dy = (3u^2 - 6u)(2x). *I'm fine so far, keep going*

    We are substituting x^2 + 2 for u in the expression for dy/dx giving
    [3(x^2+2)^2 - 6(x^2+2)](2x)

    *I understand everything above but the next step confuses me*

    6x(x^2+2)[(x^2+2)-2]
    I don't know what steps they took to arrive at that equation. Could you possible show me the 'longer' process?

    It goes on to give:
    6x(x^2+2)(x^2)=
    6x^3(x^2+2)
    First take common factor of 3 of the terms inside the square brackets outside
    the square brackets

    <br />
[3(x^2+2)^2 - 6(x^2+2)](2x) =6x [(x^2+2)^2-2(x^2+2)]<br />

    Now we also have a common factor of (x^2+2) inside the square
    brackets which we also take outside these brackets:

    <br />
[3(x^2+2)^2 - 6(x^2+2)](2x) =6x [(x^2+2)^2-2(x^2+2)]<br />
=6x (x^2+2)[(x^2+2)-2]<br />

    RonL
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  3. #3
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    Hello, Becky (and anyone listening in) . . .


    You might as well get used to simplifying messy derivatives.
    . . It comes up quite often.


    Example: Find the derivative of 2x-3)^2(x+5)^4" alt="f(x)\:=\2x-3)^2(x+5)^4" />

    Product Rule: . 2x-3)^2\cdot4(x+5)^3 + 2(2x-3)\cdot2\cdot(x+5)^4" alt="f'(x)\:=\2x-3)^2\cdot4(x+5)^3 + 2(2x-3)\cdot2\cdot(x+5)^4" />

    We have: . f'(x) \:=\:4(2x-3)^2(x+5)^3 + 4(2x-3)(x+5)^4 ...
    You'd probably stop here, right? *


    Look at the common factors: . 4,\;(2x-3)\text{, and }(x+5)^3

    Factor them out: . f'(x)\:=\:4(2x-3)(x+5)^3\,\left[(2x - 3) + (x + 5)\right]

    . . and we have: . \boxed{f'(x)\:=\:4(2x-3)(x+5)^3(3x+2)}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Your next question is (naturally) "Do we haaafta?"
    (I'm not laughing at you . . . Years ago, I asked the same question.)

    Normally, we take derivatives for a purpose, not just for exercise.

    A common problem is to find the extreme values of a function.
    We would set the derivative equal to zero and solve for x.

    With your original form: . 4(2x-3)^2(x+5)^3 + 4(2x-3)(x+5)^4 \:=\:0
    . . you might multiply it out and stare at the fifth-degree equation for days.


    But recall the procedure for solving an equation of any degree:
    . . Factor the expression, set each factor equal to zero, and solve.

    With the factored form, we have: . 4(2x-3)(x+5)^3(3x+2) \:=\:0

    . . and we can "eyeball" the solutions: . x \:=\:\frac{3}{2},\;\text{-}5,\;\text{-}\frac{2}{3}

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