Respuesta :

LammettHash
Not sure if you mean to ask for the first order partial derivatives, one wrt x and the other wrt y, or the second order partial derivative, first wrt x then wrt y. I'll assume the former.

[tex]\dfrac\partial{\partial x}(2x+3y)^{10}=10(2x+3y)^9\times2=20(2x+3y)^9[/tex]

[tex]\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9[/tex]

Or, if you actually did want the second order derivative,

[tex]\dfrac{\partial^2}{\partial y\partial x}(2x+3y)^{10}=\dfrac\partial{\partial y}\left[20(2x+3y)^9\right]=180(2x+3y)^8\times3=540(2x+3y)^8[/tex]

and in case you meant the other way around, no need to compute that, as [tex]z_{xy}=z_{yx}[/tex] by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because [tex]z[/tex] is a polynomial).

Answer:

Step-by-step explanation:

Given that z is a function of x,y

[tex]z=(2x+3y)^{10}[/tex]

Partially differentiate wrt x and y separately to get

[tex]\dfrac\partial{\partial x}(2x+3y)^{10}=10(2x+3y)^9\times2=20(2x+3y)^9\\\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9[/tex]