Re: Difference between partial derivates?

On 29-10-2007 8:03, saneman wrote:

A function f is f:R^3 -> R and is C^2.

The partial derivates are givin as:

par f/par x = y^2 + 2xz (1)
par f/par y = g(x,y,z) (2)
par f/par z = y + x^2 (3)

I am trying to find:

(par g/par x) - (par g/par z)

First I would try to find (2).

I don't know what you mean by this.

But is the relation (3)-(1) = (2) valid?

No.

Anyway,

(par g/par x) - (par g/par z) =

= (par^2 f)/((par x)(par y)) - (par^2 f)/((par z)(par y))

= (par^2 f)/((par y)(par x)) - (par^2 f)/((par y)(par z))

= 2y - 1.


Why?

Why? Why *what*? There three equalities here. Which one don't you
understand?

Actually the first line. What rule do you use to write:

(par g/par x) - (par g/par z) = (par^2 f)/((par x)(par y)) - (par^2 f)/((par z)(par y))

You now that g = (par f)/(par y), right?! So, since these two functions
are equal, their partial derivatives with respect to _x_ are also equal.
But what this means is that

(par g)/(par x) = (par ((par f)/(par y)))/(par x)

= (par^2 f)/((par x)(par y))

Best regards,

Jose Carlos Santos
.

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