What is the Partial Derivation Question in Mary Boas's Book?

[knockout_artist]

From Mary Boas's book, chapter 4.

<sub>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 ??</sub>



Thank you.

 

[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.

[tex]\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}[/tex]

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)
[tex]3{x^2}y[/tex]

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

2)[tex]y{e^{xy}}[/tex]

Which is the product of two functions of y, viz y and e^xy

This will yield two terms according to the product rule,

differentiating y with respect to -y yields 1 and so multiplied by e^xy yields the second term -e^xy

differentiating e^xy with respect to y yields xe^xy and so multiplied by -y yields -xye^xy

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

 

[knockout_artist]

It make perfec sense.
Thank you very much!