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Math Help - [SOLVED] Differentiate

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Differentiate

    I want to differentiate this 2e^{2x}(Acos(x)-Bsin(x))+e^{2x}(-Asin(x)+Bcos(x))

    I have put the a terms together to make it simpler am i right in doing so?

    Ae^{2x}(2cos(x) - sin(x)))+ Be^{2x}(2sin(x) + cos(x))

    I need to differentiate this when I try it got really messy and could not complete it.
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  2. #2
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    Quote Originally Posted by ronaldo_07 View Post
    I want to differentiate this 2e^{2x}(Acos(x)-Bsin(x))+e^{2x}(-Asin(x)+Bcos(x))

    I have put the a terms together to make it simpler am i right in doing so?

    Ae^{2x}(2cos(x) - sin(x)))+ Be^{2x}(2sin(x) + cos(x))

    I need to differentiate this when I try it got really messy and could not complete it.
    Take out a common factor of e^{2x} and you get

    f(x) = e^{2x}(2A\cos{x} - 2B\sin{x} - A\sin{x} + B\cos{x})

    f(x) = e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]


    Now differentiate using the product rule

    f'(x) = 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}] - e^{2x}[(2A + B)\sin{x} + (A + 2B)\cos{x}].
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  3. #3
    Member ronaldo_07's Avatar
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    Quote Originally Posted by Prove It View Post
    Take out a common factor of e^{2x} and you get

    f(x) = e^{2x}(2A\cos{x} - 2B\sin{x} - A\sin{x} + B\cos{x})

    f(x) = e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]


    Now differentiate using the product rule

    f'(x) = 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}] - e^{2x}[(2A + B)\sin{x} + (A + 2B)\cos{x}].
    for the first part: 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]

    Would u= 2e^{2x} and v= cos{x} - sin{x}
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  4. #4
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    Quote Originally Posted by ronaldo_07 View Post
    for the first part: 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]

    Would u= 2e^{2x} and v= cos{x} - sin{x}
    No. Let u = e^{2x} and v = everything in the square brackets.

    So the derivative is \frac{du}{dx}\times v + u\times\frac{dv}{dx}.
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  5. #5
    Member ronaldo_07's Avatar
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    I differentiated the first part 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]


    and got 2e^{2x}[(2A+B)cos(x)-(A+2B)sin(x)]-2e^{2x}[(2A+B)sin(x)+(A+2B)cos(x)]

    is this correct? then i differentiate the other part and put them together?
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  6. #6
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    Quote Originally Posted by ronaldo_07 View Post
    I differentiated the first part 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]


    and got 2e^{2x}[(2A+B)cos(x)-(A+2B)sin(x)]-2e^{2x}[(2A+B)sin(x)+(A+2B)cos(x)]

    is this correct? then i differentiate the other part and put them together?
    I've already given you the solution and answer.
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