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Please Help - For the equation given below, evaluate y′ at the point (−1,2).
ey+19−e2=3x2+4y2.


y′ at (−1,2) =

Please Help For the equation given below evaluate y at the point 12 ey19e23x24y2 y at 12 class=

Respuesta :

loneguy

[tex]e^y +19-e^2 = 3x^2 +4y^2\\\\\implies \dfrac{d}{dx}(e^y +19-e^2) = \dfrac{d}{dx}(3x^2 +4y^2)\\\\\implies e^y \dfrac{dy}{dx}+0 = 3\dfrac{d}{dx} x^2 +4\dfrac{d}{dx} y^2\\\\\implies e^y y' = 6x + 8y y'\\\\\implies e^y y' -8yy' = 6x\\\\\implies y'(e^y -8y) = 6x\\\\\implies y' = \dfrac{6x}{e^y -8y}\\\\\text{At point (-1,2)}\\\\y' = \dfrac{6(-1)}{e^2 -8(2)} = - \left(\dfrac{6}{e^2 -16}\right) =-0.25[/tex]