Respuesta :
I'm taking the liberty of editing your function v = e5xey: It should be
v = e^5x^ey, with " ^ " indicating exponentiation.
Did you mean e^(5x) or (e^5)x? I'll assume it's e^(5x).
The partial of v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.
The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
v = e^5x^ey, with " ^ " indicating exponentiation.
Did you mean e^(5x) or (e^5)x? I'll assume it's e^(5x).
The partial of v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.
The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
The second partial derivatives are [tex]\dfrac{\partial^2 v}{\partial x^2}=25e^{5x+y},\dfrac{\partial^2 v}{\partial y^2}=e^{5x+y},\dfrac{\partial^2 v}{\partial x\partial y}=5e^{5x+y}[/tex] and [tex]\dfrac{\partial^2 v}{\partial y\partial x}=5e^{5x+y}[/tex]
Given:
The given function is:
[tex]v=e^{5x}e^y[/tex]
To find:
All the second partial derivatives.
Explanation:
We have,
[tex]v=e^{5x}e^y[/tex]
[tex]v=e^{5x+y}[/tex]
First partial derivatives are:
[tex]\dfrac{\partial v}{\partial x}=5e^{5x+y}[/tex]
[tex]\dfrac{\partial v}{\partial y}=e^{5x+y}[/tex]
Second partial derivatives are:
[tex]\dfrac{\partial^2 v}{\partial x^2}=(5+0)\times 5e^{5x+y}[/tex]
[tex]\dfrac{\partial^2 v}{\partial x^2}=25e^{5x+y}[/tex]
[tex]\dfrac{\partial^2 v}{\partial y^2}=e^{5x+y}[/tex]
Similarly,
[tex]\dfrac{\partial^2 v}{\partial x\partial y}=5e^{5x+y}[/tex]
[tex]\dfrac{\partial^2 v}{\partial y\partial x}=e^{5x+y}(5+0)[/tex]
[tex]\dfrac{\partial^2 v}{\partial y\partial x}=5e^{5x+y}[/tex]
Therefore, the second partial derivatives are [tex]\dfrac{\partial^2 v}{\partial x^2}=25e^{5x+y},\dfrac{\partial^2 v}{\partial y^2}=e^{5x+y},\dfrac{\partial^2 v}{\partial x\partial y}=5e^{5x+y}[/tex] and [tex]\dfrac{\partial^2 v}{\partial y\partial x}=5e^{5x+y}[/tex]
Learn more:
https://brainly.com/question/6630548
