# help me!!! (partial derivative)

• Aug 5th 2008, 08:17 PM
ws8211
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 (Worried)

• Aug 5th 2008, 10:57 PM
Serena's Girl
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:

$\displaystyle g = f(x,y) = ax^2 + bxy + cy^2$

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

$\displaystyle \frac { \partial g} { \partial x} = a \frac {d} {dx} (x^2) + by \frac {d} {dx} (x)$

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:

$\displaystyle \frac { \partial g} { \partial x} = 2ax + by$

Similarly, ∂g/∂y is:

$\displaystyle \frac { \partial g} { \partial y} = bx + 2cy$