'second' partial derivative of a function

[BOAS]

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?

Thanks for any help you can give!

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

 

[SteamKing]

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?

Thanks for any help you can give!

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

For the exponential term, when you take the partial derivative w.r.t. x, you must apply the product rule, since you have xy * e^xy.

 

[BOAS]

For the exponential term, when you take the partial derivative w.r.t. x, you must apply the product rule, since you have xy * e^xy.

aha! In a way I'm glad it was a silly mistake like that.

Thanks :)

 

[ellipsis]

Partial derivatives seem like a big deal at first, but they're actually extremely simple. Like you said, it's just differentiation with variable constants. All the normal rules of differentiation apply.

 

[CellCoree]

Hi there,

The 'second' partial derivative of a function refers to taking the partial derivative of the partial derivative itself. In this case, we are taking the partial derivative of the function with respect to y, and then taking the partial derivative of that result with respect to x. This is what is meant by the notation $$\frac{∂}{∂x} \frac{∂z}{∂y}$$. It is essentially a shorthand way of writing $$\frac{∂}{∂x} \left(\frac{∂z}{∂y}\right)$$.

In your example, the first step is to take the partial derivative of z with respect to y, which is $$\frac{∂z}{∂y} = x^3 - xe^{xy}$$. Then, we take the partial derivative of this result with respect to x, which gives us $$\frac{∂}{∂x} \left(\frac{∂z}{∂y}\right) = \frac{∂}{∂x} [x^3 - xe^{xy}] = 3x^2 - e^{xy} - xye^{xy}$$. This is the second partial derivative of z with respect to x and y, written as $$\frac{∂^{2}z}{∂x ∂y}$$.

I hope this helps clarify things for you. As for keeping your latex in line, you can use double dollar signs before and after your equations to make sure they are displayed properly. For example, $$x^{2}$$ will display as $$x^{2}$$.

Best of luck with your studies!