Question #140047
Find the partial derivative of f(x,y,z)=x^2y+6z^3xy
1
Expert's answer
2020-10-26T19:41:51-0400

Solution:Partial derivatives:

x(x2y+6z3xy)=2yx2y1+18yz3yxln(z)\frac{\partial \:}{\partial \:x}\left(x^{2y}+6z^{3xy}\right)=2yx^{2y-1}+18yz^{3yx}\ln \left(z\right)


y(x2y+6z3xy)=2x2yln(x)+18xz3xyln(z)\frac{\partial \:}{\partial \:y}\left(x^{2y}+6z^{3xy}\right)=2x^{2y}\ln \left(x\right)+18xz^{3xy}\ln \left(z\right)


z(x2y+6z3xy)=18xyz3xy1\frac{\partial \:}{\partial \:z}\left(x^{2y}+6z^{3xy}\right)=18xyz^{3xy-1}




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