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If f(x,y) = e^(x+y) is df/dx just e^(x+y)

  1. Feb 7, 2008 #1
    if f(x,y) = e^(x+y) is df/dx just e^(x+y)
  2. jcsd
  3. Feb 7, 2008 #2
  4. Feb 7, 2008 #3
    It truly matters how you are differentiating. Make sure you take note of the fact that the original fucntion f(x,y) is in regards to two variables, but the derivative is only in regards to one (x).
  5. Feb 7, 2008 #4
    This is basically known as taking partial derivatives.
  6. Feb 9, 2008 #5
    Be careful!

    The total differential for a function of two variables is

    [tex] df = \left( \frac{\partial f}{\partial x} \right )_y dx
    + \left( \frac{\partial f}{\partial y} \right )_x dy [/tex]

    Therefore, the derivative you want is

    [tex] \frac{df}{dx} = \left( \frac{\partial f}{\partial x} \right )_y
    + \left( \frac{\partial f}{\partial y} \right )_x \frac{dy}{dx} [/tex]

    For the function you have given,

    [tex] \frac{df}{dx} = \exp (x+y) + \exp (x+y) \frac{dy}{dx} [/tex]

    So, if you hold y constant, the second term on the right is zero and the
    total derivative reduces to the partial derivative wrt x; however,
    if you wish to know how f varies along a specified
    path in the xy-plane, [tex] y=f(x) [/tex],
    then you must take the second term into account. If you wish a concrete
    example, just read any text on chemical thermodynamics.
  7. Feb 9, 2008 #6


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    pkleinod is assuming that y is a function of x. The other responders were assuming that x and y are independent variables.
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