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    f = -\frac{1}{2}x_1^2 - x_2^2 + x_1x_2 + x_1 + x_2

    \frac{\partial^2f}{\partial x_1^2} = -1
    \frac{\partial^2f}{\partial x_2^2} = -2
    \frac{\partial^2f}{\partial x_1 \partial x_2} = 1*

    I don't understand how they got the last one*. Any help appreciated
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    (Original post by Prokaryotic_crap)
    f = -\frac{1}{2}x_1^2 - x_2^2 + x_1x_2 + x_1 + x_2

    \frac{\partial^2f}{\partial x_1^2} = -1
    \frac{\partial^2f}{\partial x_2^2} = -2
    \frac{\partial^2f}{\partial x_1 \partial x_2} = 1*

    I don't understand how they got the last one*. Any help appreciated
    They differentiated w.r.t. x_1 to get: 1-x_1 + x_2 and then differentiated w.r.t. x_2 to get 1.
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    (Original post by Farhan.Hanif93)
    They differentiated w.r.t. x_1 to get: 1-x_1 + x_2 and then differentiated w.r.t. x_2 to get 1.
    oh yeah I remember now safe
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    (Original post by Farhan.Hanif93)
    They differentiated w.r.t. x_1 to get: 1-x_1 + x_2 and then differentiated w.r.t. x_2 to get 1.
    ok, so if I have a function: f = -2x_1 + 5x_2 - 2x_1^2 - 2x_2^2 + 3x_1x_2

    are these correct:

    \frac{\partial^2 f}{\partial x_1^2} = -4
    \frac{\partial^2 f}{\partial x_2^2} = -4
    \frac{\partial^2 f}{\partial x_1 \partial x_2} = 3
    ?
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    (Original post by Prokaryotic_crap)
    ok, so if I have a function: f = -2x_1 + 5x_2 - 2x_1^2 - 2x_2^2 + 3x_1x_2

    are these correct:

    \frac{\partial^2 f}{\partial x_1^2} = -4
    \frac{\partial^2 f}{\partial x_2^2} = -4
    \frac{\partial^2 f}{\partial x_1 \partial x_2} = 3
    ?
    Yes.
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    (Original post by Farhan.Hanif93)
    Yes.
    thanks
 
 
 
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