'second' partial derivative of a function

Hello,

we haven't really covered partial differentiation in my maths course yet, but it has come up a few times in mechanics where the 'grad' operator is being introduced, so i'm trying to learn about it myself. I'm looking at the partial derivatives section in "Mathematical Methods In The Physical Sciences" book by Mary Boas.

I don't understand what is being done when a 'second' partial derivative is taken. This is the example i'm looking at;

Given $$z = f(x,y) = x^{3}y - e^{xy}$$, then

$$\frac{∂z}{∂x} = 3x^{2}y - ye^{xy}$$

$$\frac{∂z}{∂y} = x^{3} - xe^{xy}$$ (I understand this, y and then x are being regarded as constants)

it goes on to say $$\frac{∂}{∂x} \frac{∂z}{∂y} = \frac{∂^{2}z}{∂x ∂y} = 3x^2 - e^{xy} - xye^{xy}$$

I don't understand what has happened here. I thought that this means taking the partial derivative of $$\frac{∂f}{∂y}$$ with respect to $$x$$, yielding $$\frac{∂}{∂x} [x^{3} - xe^{xy}] = 3x^2 - xye^{xy}$$ but this is clearly not correct as it does not account for the extra $$-e^{xy}$$ term.

What am I misunderstanding?

p.s How do I keep my pesky latex in line?

SteamKing
Staff Emeritus
Homework Helper
I don't understand what is being done when a 'second' partial derivative is taken. This is the example i'm looking at;

Given $$z = f(x,y) = x^{3}y - e^{xy}$$, then

$$\frac{∂z}{∂x} = 3x^{2}y - ye^{xy}$$

$$\frac{∂z}{∂y} = x^{3} - xe^{xy}$$ (I understand this, y and then x are being regarded as constants)

it goes on to say $$\frac{∂}{∂x} \frac{∂z}{∂y} = \frac{∂^{2}z}{∂x ∂y} = 3x^2 - e^{xy} - xye^{xy}$$

I don't understand what has happened here. I thought that this means taking the partial derivative of $$\frac{∂f}{∂y}$$ with respect to $$x$$, yielding $$\frac{∂}{∂x} [x^{3} - xe^{xy}] = 3x^2 - xye^{xy}$$ but this is clearly not correct as it does not account for the extra $$-e^{xy}$$ term.

What am I misunderstanding?