pochodne dizz: Obliczyć pochodne I i II rzędu po x i y f(x,y)=e^{−(x2+y2+2x)}

J: f'_x = e^{−(x2+y2+2x)}(−2x + 2) f'_y} −= e^{−(x2+y2+2x)}(−2y)

J: f'−X .. w nawiasie ma być: ( −2x −2)

Dziadek Mróz: f(x, y) = e^{−(x2 + y2 + 2x)} f(x, y) = e^u u = −v v = x² + y² + 2x

d		d		d
	[f(x, y)] =		[eu] = eu *		[u] = *)
dx		dx		dx

d		d		d
	[u] =		[−v] = −		[v] = **)
dx		dx		dx

d		d		d		d		d
	[v] =		[x2 + y2 + 2x] =		[x2] +		[y2] +		[2x] =
dx		dx		dx		dx		dx

	d
= 2x + y2		[1] + 2 = 2x + 2 = 2(x + 1)
	dx

**) = −(2(x + 1)) = −2(x + 1) *) = e^{−(x2 + y2 + 2x)} * (−2(x + 1)) = −2(x + 1)e^{−(x2 + y2 + 2x)}

d		d		d
	[f(x, y)] =		[eu] = eu *		[u] = *)
dy		dy		dy

d		d		d
	[u] =		[−v] = −		[v] = **)
dy		dy		dy

d		d		d		d		d
	[v] =		[x2 + y2 + 2x] =		[x2] +		[y2] +		[2x] =
dy		dy		dy		dy		dy

	d		d
= x2		[1] + 2y + 2x		[1] = 2y
	dy		dy

**) = −2y *) = e^{−(x2 + y2 + 2x)} * (−2y) = −2ye^{−(x2 + y2 + 2x)}

Dziadek Mróz: A resztę sam

dizz: dzięki wielkie teraz już z górki