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Math Help - Please explain a simple differentiation

  1. #1
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    Please explain a simple differentiation

    I saw a manipulation that goes like this:
    {{\partial E} \over {\partial A^2}} = {{\partial E} \over {2A \partial A}}

    Not familiar with this one...can anyone enlighten me?
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  2. #2
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    Quote Originally Posted by chopet View Post
    I saw a manipulation that goes like this:
    {{\partial E} \over {\partial A^2}} = {{\partial E} \over {2A \partial A}}

    Not familiar with this one...can anyone enlighten me?
    Hmmm because \partial (A^2)=2A \partial A ? it works like an usual differentiation : \frac{d}{dx}(x^2)=2x \implies d(x^2)=2xdx

    Now, if we're talking about second differentiation, we should have had \frac{\partial^{\color{red}2}E}{\partial A^2}, if E is considered as a function of A.
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  3. #3
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    Hello, chopet!
    I saw a manipulation that goes like this: . \frac{\partial E}{\partial \!A^2} \:= \:\frac{\partial E}{2A\, \partial\! A} . . . . I bet you didn't!

    Not familiar with this one...can anyone enlighten me?
    It would help if you copied it correctly!


    It actually says: . \frac{\partial^2\!E}{\partial\! A^2} \;=\;\frac{\partial^2\!E}{\partial\! A\,\partial\! A}

    Let's say we have a multivariable function: . E \;=\;f(A,B)
    . . E is function of both A and B.


    Take the partial derivative with respect to A\!:\;\;\frac{\partial E}{\partial\! A}

    Now take the partial derivative of that with respect to A\!:\;\;\frac{\partial}{\partial\! A}\!\left(\frac{\partial E}{\partial\! A}\right)

    This can be written: . \frac{\partial\!\cdot\partial E}{\partial\! A\cdot\partial\! A} \;=\;\boxed{\frac{\partial^2\!E}{\partial\! A\,\partial\! A}}\;\text{ or }\;\boxed{\frac{\partial^2\!E}{\partial\! A^2}}

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  4. #4
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    ok. I've had the equation scanned from the economics textbook dealing with Asset Pricing:

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  5. #5
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    Quote Originally Posted by Moo View Post
    Hmmm because \partial (A^2)=2A \partial A ? it works like an usual differentiation : \frac{d}{dx}(x^2)=2x \implies d(x^2)=2xdx
    couldn't see this at first.
    thanks!
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