Step-by-step explanation:
[tex]5. {x}^{3} - xy + {y}^{2} = 4[/tex]
[tex] \frac{dy}{dx} ( {x}^{3} - xy + {y}^{2} ) = \frac{dy}{dx} (4)[/tex]
[tex]3 {x}^{2} - x(1) \frac{dy}{dx} + 1(y) + 2y \frac{dy}{dx} [/tex]
Combine the dy/dx.
[tex] \frac{dy}{dx} ( - x + 2y) + y + 3 {x}^{2} [/tex]
[tex] \frac{dy}{dx} ( - x + 2y) = - 3 {x}^{2} - y[/tex]
[tex] \frac{dy}{dx} = \frac{ - 3 {x}^{2} - y}{ - x + 2y} [/tex]
[tex] \frac{3 {x}^{2} + y }{x - 2y} [/tex]
Answer:
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{y-3x^2}{2y-x}[/tex]
Step-by-step explanation:
using the product rule to differentiate - xy then
3x² - (x[tex]\frac{dy}{dx}[/tex] + y(1) ) + 2y[tex]\frac{dy}{dx}[/tex] = 0
3x² - x[tex]\frac{dy}{dx}[/tex] - y + 2y[tex]\frac{dy}{dx}[/tex] = 0
3x² + [tex]\frac{dy}{dx}[/tex] (2y - x) - y = 0 (subtract 3x² - y from both sides )
[tex]\frac{dy}{dx}[/tex] (2y - x) = y - 3x² ← divide both sides by (2y - x)
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{y-3x^2}{2y-x}[/tex]
