Partial Derivatives

Dave · 19-05-2004 07:08PM #1

Need a hand with the explanation of this. Sorry about the threads, it's exam time. Anyway here's the question.

Find all the first order and second order partial derivatives for f(x, y) = sin(xy)

daveirl · 22-05-2004 02:30AM

This post has been deleted.

silverside · 22-05-2004 03:50AM

you missed the 2 mixed partial derivatives

, which are equal (function is twice ctsly diff'able)

d2f/dydx = d2f/dxdy = cos xy - (xy sin xy)

ApeXaviour · 24-05-2004 12:33AM

lets say
df/du = df/dx + df/dy

then what would the answer to

d2f/du2 = ?

How do you mess around with these things to get what you want?

silverside · 24-05-2004 01:43AM

d2f/du2 = (d/du) of (df/du). so basically differentiate each term in the expression of df/du again. Sorry if this is not clear - you may need to look at your notes and/or an analysis book for further explantaion / examples.

ApeXaviour · 24-05-2004 02:34PM

Ok but I don't have the derivatives of them.

If I want to get it in terms of df/du, df/dx, df/dy etc.

How do I do that d'ya reckon?

ApeXaviour · 24-05-2004 05:26PM

Apologies I wan't giving enough information. I worked it out now.

Assuming f is a continious function then:

since df/du = df/dx + df/dy

then also as you said

d2f/du2 = d/du (df/dx + df/dy)

and since

d/du = d/dx + d/dy

The next line is:

d2f/du2 = (d/dx + d/dy) (df/dx + df/dy)

which when done out boils down to:

d2f/du2 = d2f/dx2 + d2f/dxdy +d2f/dydx + d2f/dx2

That's what I was looking for, but thanks, ya sent me on the right road.

Dave · 25-05-2004 05:28PM

Cheers lads. The exam isn't on till friday, and that question is worth 4% to me.

Partial Derivatives

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