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shamil98
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2wouldn't you start by finding fz first ?

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

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

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

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

hartnn
 one year 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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