Finding $frac{partial^2 u}{partial x partial y}$ of $u=f(x^2 z, z^2+x, y^2z)$.
$begingroup$
$$
A=x^2z,;B=z^2+x,;C=y^2z\
u'_x=f'_A A'_x+f'_B B'_x+f'_C C'_x=f'_Acdot 2xz+f'_B\
u''_{xy}=(f'_Acdot 2xz+f'_B)'_y=f''_{Ay}cdot 2xz+f''_{By}=2xzfrac{partial^2 f}{partial(x^2z)partial y}+frac{partial^2 f}{partial(z^2+x)partial y}=\
=2xzfrac{partial^2 f}{2xzpartial xpartial y+x^2partial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}=\
=2zfrac{partial^2 f}{2zpartial xpartial y+xpartial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}
$$
Am I right?
calculus partial-derivative
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
$endgroup$
add a comment |
$begingroup$
$$
A=x^2z,;B=z^2+x,;C=y^2z\
u'_x=f'_A A'_x+f'_B B'_x+f'_C C'_x=f'_Acdot 2xz+f'_B\
u''_{xy}=(f'_Acdot 2xz+f'_B)'_y=f''_{Ay}cdot 2xz+f''_{By}=2xzfrac{partial^2 f}{partial(x^2z)partial y}+frac{partial^2 f}{partial(z^2+x)partial y}=\
=2xzfrac{partial^2 f}{2xzpartial xpartial y+x^2partial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}=\
=2zfrac{partial^2 f}{2zpartial xpartial y+xpartial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}
$$
Am I right?
calculus partial-derivative
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
$endgroup$
$begingroup$
Leaving $partial$ x$^2$z is inadequate.
$endgroup$
– William Elliot
Dec 5 '18 at 3:19
add a comment |
$begingroup$
$$
A=x^2z,;B=z^2+x,;C=y^2z\
u'_x=f'_A A'_x+f'_B B'_x+f'_C C'_x=f'_Acdot 2xz+f'_B\
u''_{xy}=(f'_Acdot 2xz+f'_B)'_y=f''_{Ay}cdot 2xz+f''_{By}=2xzfrac{partial^2 f}{partial(x^2z)partial y}+frac{partial^2 f}{partial(z^2+x)partial y}=\
=2xzfrac{partial^2 f}{2xzpartial xpartial y+x^2partial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}=\
=2zfrac{partial^2 f}{2zpartial xpartial y+xpartial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}
$$
Am I right?
calculus partial-derivative
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
$endgroup$
$$
A=x^2z,;B=z^2+x,;C=y^2z\
u'_x=f'_A A'_x+f'_B B'_x+f'_C C'_x=f'_Acdot 2xz+f'_B\
u''_{xy}=(f'_Acdot 2xz+f'_B)'_y=f''_{Ay}cdot 2xz+f''_{By}=2xzfrac{partial^2 f}{partial(x^2z)partial y}+frac{partial^2 f}{partial(z^2+x)partial y}=\
=2xzfrac{partial^2 f}{2xzpartial xpartial y+x^2partial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}=\
=2zfrac{partial^2 f}{2zpartial xpartial y+xpartial ypartial z}+frac{partial^2 f}{2zpartial ypartial z+partial xpartial y}
$$
Am I right?
calculus partial-derivative
calculus partial-derivative
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
edited Dec 5 '18 at 23:15
guarandoo
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
asked Dec 4 '18 at 19:54
guarandooguarandoo
133
133
$begingroup$
Leaving $partial$ x$^2$z is inadequate.
$endgroup$
– William Elliot
Dec 5 '18 at 3:19
add a comment |
$begingroup$
Leaving $partial$ x$^2$z is inadequate.
$endgroup$
– William Elliot
Dec 5 '18 at 3:19
$begingroup$
Leaving $partial$ x$^2$z is inadequate.
$endgroup$
– William Elliot
Dec 5 '18 at 3:19
$begingroup$
Leaving $partial$ x$^2$z is inadequate.
$endgroup$
– William Elliot
Dec 5 '18 at 3:19
add a comment |
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$begingroup$
Leaving $partial$ x$^2$z is inadequate.
$endgroup$
– William Elliot
Dec 5 '18 at 3:19