'second' partial derivative of a function

  • Thread starter BOAS
  • Start date
  • #1
555
19
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?
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
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 * exy.
 
  • #3
555
19
For the exponential term, when you take the partial derivative w.r.t. x, you must apply the product rule, since you have xy * exy.

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

Thanks :)
 
  • #4
158
23
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.
 

Related Threads on 'second' partial derivative of a function

Replies
1
Views
909
Replies
6
Views
13K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
6
Views
998
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
6
Views
577
Replies
5
Views
2K
Replies
6
Views
360
Replies
6
Views
2K
  • Last Post
Replies
7
Views
5K
Top