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.


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

by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because

is a polynomial).
Answer:
( 2x - 1)
Step-by-step explanation:
Answer:
.......
Step-by-step explanation:
<em>whaaat</em><em>?</em><em> </em>
<em>ok</em><em> </em><em>cool</em><em> </em><em>bye</em><em> </em>
answer is (-7,2)
y = -x -5
y= x+9
Both equations have y on the left hand side
So we equate both equations
We replace -x-5 for y in the second equation
-x -5 = x+9
Subtract x on both sides
-2x -5 = 9
Now add 5 on both sides
-2x = 14
Divide by -2 from both sides
x = -7
Now plug in -7 for x in the first equation
y = -x -5
y = -(-7) -5= 7-5 = 2
So answer is (-7,2)
Answer:
4th option
Step-by-step explanation:
7x(8x² - 4y² + 3y - 5) ← multiply each term in the parenthesis by 7x
= 56x³ - 28xy² + 21xy - 35x