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shamil98
 3 years ago
\[\text{Find}~ f_{xxyzz}~ \text{for} f(x,y,z) = z^4y^3 \ln x\]
is this right :
\[f_x = \frac{ z^4y^3 }{ x }\]
\[f_{xx} = \frac{ z^4y^3 }{ x^2 }\]
\[f_{xxy} = \frac{ 3z^4y^2 }{ x^2 }\]
\[f_{xxyz} = \frac{ 12z^3y^2 }{ x^2 }\]
\[f_{xxyzz} = \frac{ 36z^2y^2 }{ x^2 }\]
shamil98
 3 years ago
\[\text{Find}~ f_{xxyzz}~ \text{for} f(x,y,z) = z^4y^3 \ln x\] is this right : \[f_x = \frac{ z^4y^3 }{ x }\] \[f_{xx} = \frac{ z^4y^3 }{ x^2 }\] \[f_{xxy} = \frac{ 3z^4y^2 }{ x^2 }\] \[f_{xxyz} = \frac{ 12z^3y^2 }{ x^2 }\] \[f_{xxyzz} = \frac{ 36z^2y^2 }{ x^2 }\]

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

shamil98
 3 years ago
Best ResponseYou've already chosen the best response.1the problem itself is asking for \[f_{xxyzz}\] the first part is \[f_x\]

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2no, dw:1388132091438:dw

shamil98
 3 years ago
Best ResponseYou've already chosen the best response.1so the first part would be: \[f_z = 4z^3y^3 \ln x\]

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