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\[\text{Find}~ f_{xxyzz}~ \text{for} f(x,y,z) = z^4y^3 \ln x\]
is this right :
\[f_x = \frac{ z^4y^3 }{ x }\]
\[f_{xx} = \frac{ z^4y^3 }{ x^2 }\]
\[f_{xxy} = \frac{ 3z^4y^2 }{ x^2 }\]
\[f_{xxyz} = \frac{ 12z^3y^2 }{ x^2 }\]
\[f_{xxyzz} = \frac{ 36z^2y^2 }{ x^2 }\]
 3 months ago
 3 months ago
\[\text{Find}~ f_{xxyzz}~ \text{for} f(x,y,z) = z^4y^3 \ln x\] is this right : \[f_x = \frac{ z^4y^3 }{ x }\] \[f_{xx} = \frac{ z^4y^3 }{ x^2 }\] \[f_{xxy} = \frac{ 3z^4y^2 }{ x^2 }\] \[f_{xxyz} = \frac{ 12z^3y^2 }{ x^2 }\] \[f_{xxyzz} = \frac{ 36z^2y^2 }{ x^2 }\]
 3 months ago
 3 months ago

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hartnnBest ResponseYou've already chosen the best response.2
wouldn't you start by finding fz first ?
 3 months ago

shamil98Best ResponseYou've already chosen the best response.1
the problem itself is asking for \[f_{xxyzz}\] the first part is \[f_x\]
 3 months ago

hartnnBest ResponseYou've already chosen the best response.2
the final answer would be same, but your method is incorrect
 3 months ago

hartnnBest ResponseYou've already chosen the best response.2
no, dw:1388132091438:dw
 3 months ago

shamil98Best ResponseYou've already chosen the best response.1
so the first part would be: \[f_z = 4z^3y^3 \ln x\]
 3 months ago

hartnnBest ResponseYou've already chosen the best response.2
yes, continue that way ....and you'll get the same final answer :)
 3 months ago
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