Respuesta :

apologiabiology
implicit differentiation
chain rule is important here

I'll show the steps partially
[tex]e^y+xe^yy'-4y'=2[/tex]
[tex]xe^yy'-4y'=2-e^y[/tex]
[tex]y'(xe^y-4)=2-e^y[/tex]
[tex]y'=\dfrac{2-e^y}{xe^y-4}[/tex]
now evaluate for (-2,2)
x=-2 and y=2
[tex]y'=\dfrac{2-e^2}{-2e^2-4}[/tex]
[tex]y'=\dfrac{2-e^2}{-2e^2-4}[/tex]
[tex]y'=\dfrac{e^2-2}{2e^2+4}[/tex]
that's it, simplest form
ernikhil

The value of [tex]y'[/tex] at [tex](-2,2)[/tex] is [tex]y'=\dfrac{2-e^2}{-2e^2-4}\\[/tex].

The given equation is [tex]xe^y-4y=2x-4-2e^2[/tex].

Differentiate the given equation with respect to [tex]x[/tex],

[tex]\dfrac{d}{dx}(xe^y-4y)=\dfrac{d}{dx}(2x-4-2e^2)\\e^y+xy'e^y-4y'=2\\y'(xe^y-4)=2-e^y\\y'=\dfrac{2-e^y}{xe^y-4}[/tex]

Substitute the given parameters as-

[tex]y'=\dfrac{2-e^y}{xe^y-4}\\y'=\dfrac{2-e^2}{-2e^2-4}\\[/tex]

Hence, the value of [tex]y'[/tex] at [tex](-2,2)[/tex] is [tex]y'=\dfrac{2-e^2}{-2e^2-4}\\[/tex].

Learn more about differentiation here:

https://brainly.com/question/24898810

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