fukncje dwoch zmiennych studentka: sprawdz czy zachodzi relacja:

	f'(x,y)		f'(x,y)
x(		−1)+y(		−1}=f(x,y)
	x'		y'

F(x,y)=(x+y)ln(x+y)

Dziadek Mróz: Mało dopracowane zadanie.

	d		d		d2
Co oznacza zapis f'(x, y)? Czy jest to		f(x, y),		f(x, y) czy		f(x, y)
	dx		dy		dxdy

ICSP: http://s10.ifotos.pl/img/kdpng_saxxwwa.png

Dziadek Mróz: f(x, y) = (x + y)ln(x + y) z = (x + y)ln(x + y) −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− z = uv u = x + y v = ln(u) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

d		d		d		d
	[z] =		[uv] =		[u]v + u		[v] = *)
dx		dx		dx		dx

d		d		d		d
	[u] =		[x + y] =		[x] +		[y] = 1 + 0 = 1
dx		dx		dx		dx

d		d		1		d		1		1
	[v] =		[ln(u)] =		*		[u] =		* 1 =
dx		dx		u		dx		x + y		x + y

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

	1
*) = 1*ln(x + y) + (x + y)		= ln(x + y) + 1
	x + y

d
	f(x, y) = ln(x + y) + 1
dx

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

d		d		d		d
	[z] =		[uv] =		[u]v + u		[v] = **)
dy		dy		dy		dy

d		d		d		d
	[u] =		[x + y] =		[x] +		[y] = 0 + 1 = 1
dy		dy		dy		dy

d		d		1		d		1		1
	[v] =		[ln(u)] =		*		[u] =		* 1 =
dy		dy		u		dy		x + y		x + y

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

	1
**) = 1*ln(x + y) + (x + y)		= ln(x + y) + 1
	x + y

d
	f(x, y) = ln(x + y) + 1
dy

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

	d		d
x(		f(x, y) − 1) + y(		f(x, y) − 1) = f(x, y)
	dx		dy

L = x(ln(x + y) + 1 − 1) + y(ln(x + y) + 1 − 1) = = xln(x + y) + yln(x + y) = ln(x + y)(x + y) = P