How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation given $x^{2} + 3 x y + y^{2} = 0$ $x^2+3xy+y^2=0$ ?

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How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation given $x^{2} + 3 x y + y^{2} = 0$ $x^2+3xy+y^2=0$ ?

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Answer 1 · 2017-06-17T18:03:13

Given: $x^{2} + 3 x y + y^{2} = 0$ $x^2+3xy+y^2=0$

Differentiate each term with respect to x:

$\frac{d (x^{2})}{d x} + \frac{3 d (x y)}{d x} + \frac{d (y^{2})}{d x} = \frac{d (0)}{d x}$ $(d(x^2))/dx + (3d(xy))/dx+ (d(y^2))/dx=(d(0))/dx$

Use the power rule, $\frac{d y}{d x} = n x^{x - 1}$ $dy/dx = nx^(x-1)$ , on the first term:

$2 x + \frac{3 d (x y)}{d x} + \frac{d (y^{2})}{d x} = \frac{d (0)}{d x}$ $2x + (3d(xy))/dx+ (d(y^2))/dx=(d(0))/dx$

Use the product rule, $\frac{d (x y)}{d x} = \frac{d x}{d x} y + x \frac{d y}{d x} = y + x \frac{d y}{d x}$ $(d(xy))/dx= dx/dxy+xdy/dx = y + xdy/dx$ on the second term:

$2 x + 3 (y + x \frac{d y}{d x}) + \frac{d (y^{2})}{d x} = \frac{d (0)}{d x}$ $2x + 3(y + xdy/dx)+ (d(y^2))/dx=(d(0))/dx$

Use the chain rule, $\frac{d (y^{2})}{d x} = 2 y \frac{d y}{d x}$ $(d(y^2))/dx=2ydy/dx$ , on the third term:

$2 x + 3 (y + x \frac{d y}{d x}) + 2 y \frac{d y}{d x} = \frac{d (0)}{d x}$ $2x + 3(y + xdy/dx)+ 2ydy/dx=(d(0))/dx$

The derivative of a constant is 0:

$2 x + 3 (y + x \frac{d y}{d x}) + 2 y \frac{d y}{d x} = 0$ $2x + 3(y + xdy/dx)+ 2ydy/dx=0$

Distribute the 3:

$2 x + 3 y + 3 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ $2x + 3y + 3xdy/dx+ 2ydy/dx=0$

Move all of the terms that do not contain $\frac{d y}{d x}$ $dy/dx$ to the right:

$3 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = - (2 x + 3 y)$ $3xdy/dx+ 2ydy/dx=-(2x+3y)$

Factor out $\frac{d y}{d x}$ $dy/dx$ :

$(3 x + 2 y) \frac{d y}{d x} = - (2 x + 3 y)$ $(3x+2y)dy/dx=-(2x+3y)$

Divide by $3 x + 2 y$ $3x+2y$ :

$\frac{d y}{d x} = - \frac{2 x + 3 y}{3 x + 2 y}$ $dy/dx=-(2x+3y)/(3x+2y)$

Answer 2 · 2017-06-17T18:43:14

$\frac{d y}{d x} = - \frac{2 x + 3 y}{3 x + 2 y}$ $dy/dx=-(2x+3y)/(3x+2y)$

Explanation:

$differentiate implicitly with respect to x$ $"differentiate "color(blue)"implicitly with respect to x"$

$the term 3 x y is differentiated using the product rule$ $"the term " 3xy" is differentiated using the "color(blue)"product rule"$

#rArr2x+3(x.dy/dx+y.1)+2y.dy/dx=0#

$\Rightarrow 2 x + 3 x \frac{d y}{d x} + 3 y + 2 y \frac{d y}{d x} = 0$ $rArr2x+3xdy/dx+3y+2ydy/dx=0$

$\Rightarrow \frac{d y}{d x} (3 x + 2 y) = - 2 x - 3 y$ $rArrdy/dx(3x+2y)=-2x-3y$

$\Rightarrow \frac{d y}{d x} = - \frac{2 x + 3 y}{3 x + 2 y}$ $rArrdy/dx=-(2x+3y)/(3x+2y)$

Answer 3 · 2017-06-17T20:49:17

$\frac{d y}{d x} = - \frac{1}{2} (\sqrt{5} \pm 3)$ $(dy)/(dx)=-1/2(sqrt 5 pm 3)$

Explanation:

From $x^{2} + 3 x y + y^{2} = 0 \to y = - \frac{1}{2} (\sqrt{5} \pm 3) x$ $x^2 + 3 x y + y^2=0 -> y = -1/2(sqrt 5 pm 3)x$

then

$\frac{d y}{d x} = - \frac{1}{2} (\sqrt{5} \pm 3)$ $(dy)/(dx)=-1/2(sqrt 5 pm 3)$

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How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation given $x^{2} + 3 x y + y^{2} = 0$ $x^2+3xy+y^2=0$ ?

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