Matematické Fórum

krecek · 20. 02. 2012 09:31

prosím, poradte, je tohle dobře?

$f(x, y) = sin(x^{2}y + 3y - 4x +\frac{\Pi }{3})$
$f'(x) = (0,0)$
$f'(y) = (0,0)$

moje reseni:

$f'(x) = cos (x^{2}y + 3y - 4x +\frac{\Pi }{3}) \cdot (2xy + 0 -4 + 0) = cos (x^{2}y + 3y - 4x +\frac{\Pi }{3}) \cdot (2xy -4)$
$f'x(0,0) = cos (0^{2}\cdot 0 + 3\cdot 0 - 4\cdot 0 +\frac{\Pi }{3}) \cdot (2\cdot 0\cdot 0 -4 ) = cos (\frac{\Pi }{3}) \cdot (-4) = \underline{\text{-2}}$

$f'(y) = cos (x^{2}y + 3y - 4x +\frac{\Pi }{3}) \cdot (x^2 + 3 -0 + 0) = cos (x^{2}y + 3y - 4x +\frac{\Pi }{3}) \cdot (x^2 + 3)$
$f'y(0,0) = cos (0^{2}\cdot 0 + 3\cdot 0 - 4\cdot 0 +\frac{\Pi }{3}) \cdot (0^2 + 3) = cos (\frac{\Pi }{3}) \cdot (3) = \underline{{\frac{3}{2}}}$