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Math Help - help me!!! (partial derivative)

  1. #1
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    help me!!! (partial derivative)

    1. Take the derivatives of the following functions with respect to x (∂/∂x):

    2. Calculate the partial derivative with respect to x (∂/∂x) and y (∂/∂y ):

    3. Let g = f (x, y) = ax2 + bxy + cy2 . Calculate the partial derivative with respect to x
    (∂g/∂x) and y (∂g/∂y).

    4. Let g = f (x, y, z) = xa yb zc . Calculate the partial derivative with respect to x (∂g/∂x),
    y (∂g/∂y) and z (∂g/∂y).

    5. Let g = f (x, y) = a ln x + bln y . Calculate the partial derivative with respect to x
    (∂g/∂x) and y (∂g/∂y).

    6. Let g = f (x, y) = eax+by . Calculate the partial derivative with respect to x (∂g/∂x) and y
    (∂g/∂y).

    i do not exactly understand how to solve partial derivative

    please help me thanks!
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  2. #2
    Junior Member Serena's Girl's Avatar
    Joined
    Jul 2008
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    Smile Partial derivatives

    Let's say we are given g = f(x,y,z), and we wish to take the partial derivative of g with respect to x (i.e. ∂g/∂x).

    This simply means that we will obtain the derivative of g with respect to x while treating the other variables (i.e. y and z) as though they were constants. We will solve #3 to demonstrate:

    <br />
g = f(x,y) = ax^2 + bxy + cy^2<br />

    Solve for ∂g/∂x. We treat the variable y as though it were a constant. So...

    <br />
\frac { \partial g} { \partial x} = a \frac {d} {dx} (x^2) + by \frac {d} {dx} (x)<br />

    The third term "disappeared" because, in this case, cy^2 is treated as a constant, and the derivative of a constant is 0.

    Thus, we obtain:

    <br />
\frac { \partial g} { \partial x} = 2ax + by<br />

    Similarly, ∂g/∂y is:

    <br />
\frac { \partial g} { \partial y} = bx + 2cy<br />
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