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    So on the Jan 2012 paper for Question 7a: http://pmt.physicsandmathstutor.com/...20C3%20AQA.pdf

    I worked out dy/dx to be (1/4)x^2*e^-(1/4)x + 2x*e^-(1/2)x = 0

    What I originally done was divide through by x - why is this not allowed? So I think I basically lost an x term. And so do I have to basically remember that whenever Im finding stationary points I can't divide through by x? Does this apply normally as well?
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    (Original post by MrToodles4)
    So on the Jan 2012 paper for Question 7a: http://pmt.physicsandmathstutor.com/...20C3%20AQA.pdf

    I worked out dy/dx to be (1/4)x^2*e^-(1/4)x + 2x*e^-(1/2)x = 0
    You've got two errors in that derivative.

    \displaystyle \frac{dy}{dx}= -\frac{x^2}{4}e^{-(1/4)x}+2xe^{-(1/4)x}

    What I originally done was divide through by x - why is this not allowed? So I think I basically lost an x term. And so do I have to basically remember that whenever Im finding stationary points I can't divide through by x? Does this apply normally as well?
    If you're trying to solve an equation equal to 0, you can only divide through by a term you know to be non-zero.

    e.g xe^x=0

    You can divide by e^x since this term can never be zero.

    You cannot divide by x, since x=0 is a solution, and you'd lose that.
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    (Original post by ghostwalker)
    You've got two errors in that derivative.

    Sorry I typed in wrong.

    \displaystyle \frac{dy}{dx}= -\frac{x^2}{4}e^{-(1/4)x}+2xe^{-(1/4)x}



    If you're trying to solve an equation equal to 0, you can only divide through by a term you know to be non-zero.

    e.g xe^x=0

    You can divide by e^x since this term can never be zero.

    You cannot divide by x, since x=0 is a solution, and you'd lose that.
    Alright, that makes a lot of sense. Thank you so much I really appreciate it
 
 
 
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