Question

For the equation given below, evaluate dy dx at the point (0,2) . 2y^3 +y^2 −3x^2 =20

Answer 1

\[\frac{d(2y^3)}{dx} +\frac{d(y^2)}{dx} −\frac{d(3x^2)}{dx} =\frac{d(20)}{dx} \]

Answer 2

\[\frac{dy}{dx}6y^2+\frac{dy}{dx}2y-\frac{dx}{dx}6x=\frac{dC}{dx}0\]

Answer 3

dx/dx = 1; and dC/dx = 1 as well; prolly shoulda made that a dx/dx since the variable could have been x^0 :)

Answer 4

\[\frac{dy}{dx}(6y^2+2y)-6x=0\]
\[\frac{dy}{dx}=\frac{6x}{6y^2+2y}\]

Answer 5

and since (x=0,y=2) ; we get 6(0) = 0 for the top and that makes dy/dx = 0