I need help with all of this
1 answer:
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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).
Or, if you actually did want the second order derivative,
and in case you meant the other way around, no need to compute that, as
Answer:
x = -6 and x = 4
Step-by-step explanation:
In math, the critical points of a function are the points where the derivative equals zero.
So, first we will find the derivative of the function. The derivative is:
Now, we are going to make the derivative equal zero and find the answers of the equation.
So we have that the critical points are the answers to this equation:
and
Thus, the critical points are x=-6 and x=4