Question

What is partial derivative of z=(2x+3y)^10 with respect to x,y?

Answer 1

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.

$\dfrac\partial{\partial x}(2x+3y)^{10}=10(2x+3y)^9\times2=20(2x+3y)^9$

$\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9$

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

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

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

Answer 2

Answer:

Step-by-step explanation:

Given that z is a function of x,y

$z=(2x+3y)^{10}$

Partially differentiate wrt x and y separately to get

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