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Math Help - Finding dy/dx of equation using both chain rule and product rule

  1. #1
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    Finding dy/dx of equation using both chain rule and product rule

    Hello everyone,

    Recently I've been learning about the chain rule and the product rule, which by themselves is fairly straightforward to solve. However, it becomes a little more complex when attempting to solve an equation with both of them together. So any help in solving the following equation using the chain rule and product rule will be greatly appreciated.

    Here is the equation:

    y = (3x + 5) ^2 x (2x - 2)

    I solved the chain rule part (I think), which is the (3x + 5) ^2 segment. Using the chain rule, I got the answer 6(3x + 5), leaving me with the equation of:

    y = 6(3x + 5)(2x - 2) to be solved using the product rule. But I am not entirely confident with the steps involved, as I believe it involves factorizing and such.

    Thanks!

    Nathaniel
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  2. #2
    MHF Contributor Unknown008's Avatar
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    That's not how it is done...

    y = 6(3x + 5)(2x - 2) is wrong, because y = (3x + 5)^2 . (2x - 2) [use . if necessary to show product]

    Using product rule first:

    Let us break it up.

    Let u = (3x+5)^2, let v = (2x-2)

    then, we get:

    \dfrac{du}{dx} = 6(3x + 5) = 18x + 30

    \dfrac{dv}{dx} = 2

    Then, y = uv

    y' = udv/dx + vdu/dx

    y'= (3x+5)^2 \cdot 2 + (2x-2) \cdot (18x + 30)

    Now, you simplify:

    y'= 2(3x+5)^2 + (2x-2)(18x + 30)

    You can expand and simplify further itf you want.
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  3. #3
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    \displaystyle y = (3x + 5)^2(2x - 2)

    \displaystyle \frac{dy}{dx} = (3x + 5)^2\,\frac{d}{dx}(2x - 2) + (2x -2)\,\frac{d}{dx}[(3x + 5)^2]

    \displaystyle = 2(3x + 5)^2 + 6(3x + 5)(2x - 2)

    \displaystyle = (3x + 5)[2(3x + 5) + 6(2x - 2)]

    \displaystyle = (3x + 5)(6x + 10 + 12x - 12)

    \displaystyle = (3x + 5)(18x - 2)

    \displaystyle = 2(3x + 5)(9x - 1).
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  4. #4
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    Thank you both so much for the speedy replies, Unknown008, and Prove It. Okay, so I get both methods, but there is just one thing that I think I need clarification on:

    Prove It, can you please explain the reasoning behind the factorizing. As in I do not understand why the 6 suddenly moved in front of the (2x + 2), and why the (3x + 5) is out of the front of what seems to be a factorized equation.

    Once again, thanks so much for all the help!


    EDIT:

    Actually, I get it now, I think!

    That factorizing is just like

    x ( 2x + 6y) and when you expand you get 2x ^ 2 + 6xy

    And subbing the values of x and y, I get: 2(3x + 5) ^ 2 + 6(3x + 5)(2x - 2)

    I think that is correct reasoning. Please do clarify if I am wrong though.
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Yes, that's it

    Once you get the hang of it, you can differentiate directly to what Prove It posted.
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  6. #6
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    Actually, there's just one last thing.

    The second last line of Prove It's method reads: (3x + 5)(18x - 2)

    And the final answer is: 2(3x + 5)(9x -1)

    I'm just wondering if I need to use the product rule to achieve that final answer?

    Thanks again.

    Nathaniel.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    The conversion from the before last line to the last line is simply the factorisation of (18x -2) = 2(9x -1)

    The product rule is used only on the line where dy/dx appeared first.
    Last edited by Unknown008; November 9th 2010 at 01:34 AM. Reason: Typo... put dy/dy instead of dy/dx :o
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  8. #8
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    Very good indeed, Unknown008.

    So the equation would still be correct if I wrote it as (3x + 5) . 2(9x -1) ? It was just confusing, because the nature of (3x + 5) didn't change with the 2 stuck out the front.
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  9. #9
    MHF Contributor Unknown008's Avatar
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    It's algebraically the same thing. That's why in my post, I told you that you could simplify further if you want.

    Sometimes, for the problem, it's better to simplify for the other parts to be easier.

    Sometimes, it's just a matter of substituting values of x and/or y. Then, simplification doesn't necessaily makes it easier than before.
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  10. #10
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    Excellent. I think my troubles regarding this problem have been solved now. I'll just do some practice with similar problems. This has been a great help, and yet again a wonderful resource.

    Cheers!
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