Find dx/dy x^2y-3x^2-4=0

$x^{2} y - 3 x^{2} - 4 = 0$

Differentiate both sides of the equation.

$\frac{d}{d y} (x^{2} y - 3 x^{2} - 4) = \frac{d}{d y} (0)$

Differentiate the left side of the equation.

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By the Sum Rule, the derivative of $x^{2} y - 3 x^{2} - 4$ with respect to $y$ is $\frac{d}{d y} [x^{2} y] + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$ .

$\frac{d}{d y} [x^{2} y] + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Evaluate $\frac{d}{d y} [x^{2} y]$ .

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Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = x^{2}$ and $g (y) = y$ .

$x^{2} \frac{d}{d y} [y] + y \frac{d}{d y} [x^{2}] + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$x^{2} \cdot 1 + y \frac{d}{d y} [x^{2}] + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{2}$ and $g (y) = x$ .

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To apply the Chain Rule, set $u_{1}$ as $x$ .

$x^{2} \cdot 1 + y (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

$x^{2} \cdot 1 + y (2 u_{1} \frac{d}{d y} [x]) + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Replace all occurrences of $u_{1}$ with $x$ .

$x^{2} \cdot 1 + y (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Rewrite $\frac{d}{d y} [x]$ as $\frac{d}{d y} [x]$ .

$x^{2} \cdot 1 + y (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Multiply $x^{2}$ by $1$ .

$x^{2} + y (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Move $2$ to the left of $y$ .

$x^{2} + 2 y x \frac{d}{d y} [x] + \frac{d}{d y} [- 3 x^{2}] + \frac{d}{d y} [- 4]$

Evaluate $\frac{d}{d y} [- 3 x^{2}]$ .

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Since $- 3$ is constant with respect to $y$ , the derivative of $- 3 x^{2}$ with respect to $y$ is $- 3 \frac{d}{d y} [x^{2}]$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 3 \frac{d}{d y} [x^{2}] + \frac{d}{d y} [- 4]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{2}$ and $g (y) = x$ .

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To apply the Chain Rule, set $u_{2}$ as $x$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 3 (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [- 4]$

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 2$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 3 (2 u_{2} \frac{d}{d y} [x]) + \frac{d}{d y} [- 4]$

Replace all occurrences of $u_{2}$ with $x$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 3 (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [- 4]$

Rewrite $\frac{d}{d y} [x]$ as $\frac{d}{d y} [x]$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 3 (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [- 4]$

Multiply $2$ by $- 3$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 6 x \frac{d}{d y} [x] + \frac{d}{d y} [- 4]$

Since $- 4$ is constant with respect to $y$ , the derivative of $- 4$ with respect to $y$ is $0$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 6 x \frac{d}{d y} [x] + 0$

Simplify.

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Add $x^{2} + 2 y x \frac{d}{d y} [x] - 6 x \frac{d}{d y} [x]$ and $0$ .

$x^{2} + 2 y x \frac{d}{d y} [x] - 6 x \frac{d}{d y} [x]$

Reorder terms.

$x^{2} + 2 x y \frac{d}{d y} [x] - 6 x \frac{d}{d y} [x]$

Since $0$ is constant with respect to $y$ , the derivative of $0$ with respect to $y$ is $0$ .

$0$

Reform the equation by setting the left side equal to the right side.

$x^{2} + 2 x y x' - 6 x x' = 0$

Solve for $x'$ .

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Subtract $x^{2}$ from both sides of the equation.

$2 x y x' - 6 x x' = - x^{2}$

Factor $2 x x'$ out of $2 x y x' - 6 x x'$ .

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Factor $2 x x'$ out of $2 x y x'$ .

$2 x x' (y) - 6 x x' = - x^{2}$

Factor $2 x x'$ out of $- 6 x x'$ .

$2 x x' (y) + 2 x x' (- 3) = - x^{2}$

Factor $2 x x'$ out of $2 x x' y + 2 x x' \cdot - 3$ .

$2 x x' (y - 3) = - x^{2}$

Divide each term by $2 x (y - 3)$ and simplify.

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Divide each term in $2 x x' (y - 3) = - x^{2}$ by $2 x (y - 3)$ .

$\frac{2 x x' (y - 3)}{2 x (y - 3)} = \frac{- x^{2}}{2 x (y - 3)}$

Simplify $\frac{2 x x' (y - 3)}{2 x (y - 3)}$ .

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x x' (y - 3)}{2 x (y - 3)} = \frac{- x^{2}}{2 x (y - 3)}$

Rewrite the expression.

$\frac{x x' (y - 3)}{x (y - 3)} = \frac{- x^{2}}{2 x (y - 3)}$

Cancel the common factor of $x$ .

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Cancel the common factor.

$\frac{x x' (y - 3)}{x (y - 3)} = \frac{- x^{2}}{2 x (y - 3)}$

Rewrite the expression.

$\frac{x' (y - 3)}{y - 3} = \frac{- x^{2}}{2 x (y - 3)}$

Cancel the common factor of $y - 3$ .

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Cancel the common factor.

$\frac{x' (y - 3)}{y - 3} = \frac{- x^{2}}{2 x (y - 3)}$

Divide $x'$ by $1$ .

$x' = \frac{- x^{2}}{2 x (y - 3)}$

Simplify $\frac{- x^{2}}{2 x (y - 3)}$ .

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Cancel the common factor of $x^{2}$ and $x$ .

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Factor $x$ out of $- x^{2}$ .

$x' = \frac{x (- x)}{2 x (y - 3)}$

Cancel the common factors.

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Factor $x$ out of $2 x (y - 3)$ .

$x' = \frac{x (- x)}{x (2 (y - 3))}$

Cancel the common factor.

$x' = \frac{x (- x)}{x (2 (y - 3))}$

Rewrite the expression.

$x' = \frac{- x}{2 (y - 3)}$

Move the negative in front of the fraction.

$x' = - \frac{x}{2 (y - 3)}$

Replace $x'$ with $\frac{d x}{d y}$ .

$\frac{d x}{d y} = - \frac{x}{2 (y - 3)}$

Find dx/dy x^2y-3x^2-4=0

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