# Partial derivation question

1. Feb 8, 2012

### knockout_artist

From Mary Boas's book, chapter 4.

Z = x^3 -e^xy

1- Z(x) = 3x^2 y - ye^xy make sense according to derivation rule d(e^u)/dx = e^u.u`
2-Z(y) = 3^x - xe^xy make sense too.

3-Z(y)x)) = 3x^2 - e^xy - xye^xy where e^xy came from ??

Thank you.

2. Feb 9, 2012

### Studiot

Even though this is one of the first examples in chapter 4 a more complete reference would have helped get you a quicker response.

$$\begin{array}{l} \frac{{\partial f}}{{\partial x}} = 3{x^2}y - y{e^{xy}} \\ \frac{{\partial f}}{{\partial x\partial y}} = 3{x^2} - {e^{xy}} - xy{e^{xy}} \\ \end{array}$$

So you understand how to take the first partial differential with respect to x or y.

I have shown the one with respect to x.

The second partial that you are having trouble with is formed by taking either the first differential with respect to x and differentiating it with respect to y or the other way round. The result is the same.

I have chosen to take the first partial with respect to x and then differentiate it with respect to y.

The first partial is formed from two terms

1)
$$3{x^2}y$$

from which we get 3x2 since x is considered constant in this second differentiation.

2)$$y{e^{xy}}$$

Which is the product of two functions of y, viz y and exy

This will yield two terms according to the product rule,

differentiating y with respect to -y yields 1 and so multiplied by exy yields the second term -exy

differentiating exy with respect to y yields xexy and so multiplied by -y yields -xyexy

which is the third term you are having trouble with. does this help?

3. Feb 10, 2012

### knockout_artist

It make perfec sense.

Thank you very much!!