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).
Symmetric property, I remembered the answer for this for my summer school. This is one of the equivalence properties of equality.
Step 2: commutative property of multiplication
Step 3: multiplicative inverse
Step 4: multiplicative identity
If you would like to solve the system of equations, you can do this using the following steps:
4x^2 + 9y^2 = 72
2x - y = 4 ... 2x - 4 = y
_________
<span>4x^2 + 9y^2 = 72
</span><span>4x^2 + 9 * (2x - 4)^2 = 72
</span>4x^2 + 9 * (4x^2 - 16x + 16) = 72
4x^2 + 36x^2 - 144x + 144 = 72
40x^2 - 144x + 144 - 72 = 0
40x^2 - 144x + 72 = 0
10x^2 - 36x + 18 = 0
5x^2 - 18x + 9 = 0
(5x - 3) * (x - 3) = 0
1. 5x - 3 = 0 ... 5x = 3 ... x = 3/5
2. x = 3
<span>1. y = 2x - 4 = 2 * 3/5 - 4 = 6/5 - 20/5 = -14/5
2. y = 2x - 4 = 2 * 3 - 4 = 6 - 4 = 2
1. (x, y) = (3/5, -14/5)
2. (x, y) = (3, 2)
The correct result would be </span>(3/5, -14/5) and <span>(3, 2).</span>
Energy is the correct answer
