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

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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

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

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

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 $g (y) = y$ .

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

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

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

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 $g (y) = x$ .

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

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

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

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

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

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

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

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

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

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

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

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

Evaluate $\frac{d}{d y} [- 4 x^{2}]$ .

Since $- 4$ is constant with respect to $y$ , the derivative of $- 4 x^{2}$ with respect to $y$ is $- 4 \frac{d}{d y} [x^{2}]$ .

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

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 $g (y) = x$ .

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

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

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

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

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

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

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

$x^{2} + 2 y x x' - 4 (2 x x')$

Multiply $2$ by $- 4$ .

$x^{2} + 2 y x x' - 8 x x'$

Reorder terms.

$x^{2} + 2 x y x' - 8 x x'$

Step 3

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

$0$

Step 4

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

$x^{2} + 2 x y x' - 8 x x' = 0$

Step 5

Solve for $x'$ .

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

$2 x y x' - 8 x x' = - x^{2}$

Factor $2 x x'$ out of $2 x y x' - 8 x x'$ .

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

$2 x x' y - 8 x x' = - x^{2}$

Factor $2 x x'$ out of $- 8 x x'$ .

$2 x x' y + 2 x x' \cdot - 4 = - x^{2}$

Factor $2 x x'$ out of $2 x x' y + 2 x x' \cdot - 4$ .

$2 x x' (y - 4) = - x^{2}$

Divide each term in $2 x x' (y - 4) = - x^{2}$ by $2 x (y - 4)$ and simplify.

Divide each term in $2 x x' (y - 4) = - x^{2}$ by $2 x (y - 4)$ .

$\frac{2 x x' (y - 4)}{2 x (y - 4)} = \frac{- x^{2}}{2 x (y - 4)}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x x' (y - 4)}{2 x (y - 4)} = \frac{- x^{2}}{2 x (y - 4)}$

Rewrite the expression.

$\frac{x x' (y - 4)}{x (y - 4)} = \frac{- x^{2}}{2 x (y - 4)}$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x x' (y - 4)}{x (y - 4)} = \frac{- x^{2}}{2 x (y - 4)}$

Rewrite the expression.

$\frac{x' (y - 4)}{y - 4} = \frac{- x^{2}}{2 x (y - 4)}$

Cancel the common factor of $y - 4$ .

Cancel the common factor.

$\frac{x' (y - 4)}{y - 4} = \frac{- x^{2}}{2 x (y - 4)}$

Divide $x'$ by $1$ .

$x' = \frac{- x^{2}}{2 x (y - 4)}$

Simplify the right side.

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

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

$x' = \frac{x (- x)}{2 x (y - 4)}$

Cancel the common factors.

Factor $x$ out of $2 x (y - 4)$ .

$x' = \frac{x (- x)}{x (2 (y - 4))}$

Cancel the common factor.

$x' = \frac{x (- x)}{x (2 (y - 4))}$

Rewrite the expression.

$x' = \frac{- x}{2 (y - 4)}$

Move the negative in front of the fraction.

$x' = - \frac{x}{2 (y - 4)}$

Step 6

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

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

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

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