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](https://tex.z-dn.net/?f=%5Cdfrac%5Cpartial%7B%5Cpartial%20x%7D%282x%2B3y%29%5E%7B10%7D%3D10%282x%2B3y%29%5E9%5Ctimes2%3D20%282x%2B3y%29%5E9)
![\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9](https://tex.z-dn.net/?f=%5Cdfrac%5Cpartial%7B%5Cpartial%20y%7D%282x%2B3y%29%5E%7B10%7D%3D10%282x%2B3y%29%5E9%5Ctimes3%3D30%282x%2B3y%29%5E9)
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](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%5E2%7D%7B%5Cpartial%20y%5Cpartial%20x%7D%282x%2B3y%29%5E%7B10%7D%3D%5Cdfrac%5Cpartial%7B%5Cpartial%20y%7D%5Cleft%5B20%282x%2B3y%29%5E9%5Cright%5D%3D180%282x%2B3y%29%5E8%5Ctimes3%3D540%282x%2B3y%29%5E8)
and in case you meant the other way around, no need to compute that, as
![z_{xy}=z_{yx}](https://tex.z-dn.net/?f=z_%7Bxy%7D%3Dz_%7Byx%7D)
by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because
![z](https://tex.z-dn.net/?f=z)
is a polynomial).
Step-by-step explanation:
Answer:
slope = 3
Step-by-step explanation:
to find the slope you do (y2-y1)/(x2-x1)=m
so we can plug the points into this formula
(5,7) (2,-2)
x1 y1 x2 y2
-2-7/2-5=m
-9/-3=m
Simplifies to
3=m
It represents simulations