x-2-xy-y-2-3y-10-find-the-value-of-dy-dx-at-x-2-

Question Number 99357 by bobhans last updated on 20/Jun/20

x^{2} + xy + y^{2} - 3 y = 10, find the value of

\frac{dy}{dx} at x = 2

Commented by bemath last updated on 20/Jun/20

x = 2 \Rightarrow 4 + 2 y + y^{2} - 3 y - 10 = 0

y^{2} - y - 6 = 0 \Rightarrow {\begin{cases} y = 3 \\ y = - 2 \end{cases}

implicit differentiate

2 x + y + x \frac{dy}{dx} + 2 y \frac{dy}{dx} - 3 \frac{dy}{dx} = 0

2 x + y + (x - 3 + 2 y) \frac{dy}{dx} = 0

for (2, 3) \Rightarrow 4 + 3 + (2 - 3 + 6) \frac{dy}{dx} = 0

\frac{dy}{dx} = - \frac{7}{5}

for (2, - 2) \Rightarrow 4 - 2 + (2 - 3 - 4) \frac{dy}{dx} = 0

\frac{dy}{dx} = \frac{- 2}{(- 5)} = \frac{2}{5}

Answered by mathmax by abdo last updated on 20/Jun/20

x^{2} + y^{2} + xy - 3 y = 10 \Rightarrow y^{2} + (x - 3) y + x^{2} - 10 = 0

Δ = {(x - 3)}^{2} - 4 (x^{2} - 10) = x^{2} - 6 x + 9 - {4 x}^{2} + 40 = - {3 x}^{2} - 6 x + 49

for Δ ⩾ 0 we get y_{1} = \frac{3 - x + \sqrt{- {3 x}^{2} - 6 x + 49}}{2}

and y_{2} = \frac{3 - x - \sqrt{- {3 x}^{2} - 6 x + 49}}{2}

y = y_{1} \Rightarrow y^{^{'}} = - \frac{1}{2} + \frac{1}{2} \times \frac{- 6 x - 6}{2 \sqrt{- {3 x}^{2} - 6 x + 49}} = - \frac{1}{2} - \frac{3 x + 3}{2 \sqrt{- {3 x}^{2} - 6 x + 49}}

y^{^{'}} (2) = - \frac{1}{2} - \frac{9}{2 \sqrt{- 12 - 12 + 49}} = - \frac{1}{2} - \frac{9}{10} = \frac{- 5 - 9}{10} = \frac{- 14}{10} = - \frac{7}{5}

y = y_{2} \Rightarrow y^{^{'}} = - \frac{1}{2} - \frac{1}{2} \times \frac{- 6 x - 6}{2 \sqrt{- {3 x}^{2} - 6 x + 49}} = - \frac{1}{2} + \frac{1}{2} \frac{3 x + 3}{\sqrt{- {3 x}^{2} - 6 x + 49}} \Rightarrow

y^{^{'}} (2) = - \frac{1}{2} + \frac{9}{2 \times 5} = - \frac{1}{2} + \frac{9}{10} = \frac{- 5 + 9}{10} = \frac{4}{10} = \frac{2}{5}