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

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

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

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

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{4}$ and $g (y) = x$ .

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

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

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 4$ .

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

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

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

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

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

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

Since $- 1$ is constant with respect to $y$ , the derivative of $- x^{3} y^{2}$ with respect to $y$ is $- \frac{d}{d y} [x^{3} y^{2}]$ .

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

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 $f (y) = x^{3}$ and $g (y) = y^{2}$ .

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

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

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

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{3}$ and $g (y) = x$ .

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

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

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

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

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

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

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

$2 + x' + 4 x^{3} x' - (x^{3} (2 y) + y^{2} (3 x^{2} x'))$

Move $2$ to the left of $x^{3}$ .

$2 + x' + 4 x^{3} x' - (2 \cdot x^{3} y + y^{2} (3 x^{2} x'))$

Move $3$ to the left of $y^{2}$ .

$2 + x' + 4 x^{3} x' - (2 x^{3} y + 3 y^{2} x^{2} x')$

Simplify.

Apply the distributive property.

$2 + x' + 4 x^{3} x' - (2 x^{3} y) - (3 y^{2} x^{2} x')$

Combine terms.

Multiply $2$ by $- 1$ .

$2 + x' + 4 x^{3} x' - 2 (x^{3} y) - (3 y^{2} x^{2} x')$

Multiply $3$ by $- 1$ .

$2 + x' + 4 x^{3} x' - 2 x^{3} y - 3 y^{2} x^{2} x'$

Reorder terms.

$4 x^{3} x' - 2 x^{3} y - 3 x^{2} y^{2} x' + 2 + x'$

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.

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

Step 5

Solve for $x'$ .

Move all terms not containing $x'$ to the right side of the equation.

Add $2 x^{3} y$ to both sides of the equation.

$4 x^{3} x' - 3 x^{2} y^{2} x' + 2 + x' = 2 x^{3} y$

Subtract $2$ from both sides of the equation.

$4 x^{3} x' - 3 x^{2} y^{2} x' + x' = 2 x^{3} y - 2$

Factor $x'$ out of $4 x^{3} x' - 3 x^{2} y^{2} x' + x'$ .

Factor $x'$ out of $4 x^{3} x'$ .

$x' (4 x^{3}) - 3 x^{2} y^{2} x' + x' = 2 x^{3} y - 2$

Factor $x'$ out of $- 3 x^{2} y^{2} x'$ .

$x' (4 x^{3}) + x' (- 3 x^{2} y^{2}) + x' = 2 x^{3} y - 2$

Raise $x'$ to the power of $1$ .

$x' (4 x^{3}) + x' (- 3 x^{2} y^{2}) + x' = 2 x^{3} y - 2$

Factor $x'$ out of ${x'}^{1}$ .

$x' (4 x^{3}) + x' (- 3 x^{2} y^{2}) + x' \cdot 1 = 2 x^{3} y - 2$

Factor $x'$ out of $x' (4 x^{3}) + x' (- 3 x^{2} y^{2})$ .

$x' (4 x^{3} - 3 x^{2} y^{2}) + x' \cdot 1 = 2 x^{3} y - 2$

Factor $x'$ out of $x' (4 x^{3} - 3 x^{2} y^{2}) + x' \cdot 1$ .

$x' (4 x^{3} - 3 x^{2} y^{2} + 1) = 2 x^{3} y - 2$

Divide each term in $x' (4 x^{3} - 3 x^{2} y^{2} + 1) = 2 x^{3} y - 2$ by $4 x^{3} - 3 x^{2} y^{2} + 1$ and simplify.

Divide each term in $x' (4 x^{3} - 3 x^{2} y^{2} + 1) = 2 x^{3} y - 2$ by $4 x^{3} - 3 x^{2} y^{2} + 1$ .

$\frac{x' (4 x^{3} - 3 x^{2} y^{2} + 1)}{4 x^{3} - 3 x^{2} y^{2} + 1} = \frac{2 x^{3} y}{4 x^{3} - 3 x^{2} y^{2} + 1} + \frac{- 2}{4 x^{3} - 3 x^{2} y^{2} + 1}$

Simplify the left side.

Cancel the common factor of $4 x^{3} - 3 x^{2} y^{2} + 1$ .

Cancel the common factor.

$\frac{x' (4 x^{3} - 3 x^{2} y^{2} + 1)}{4 x^{3} - 3 x^{2} y^{2} + 1} = \frac{2 x^{3} y}{4 x^{3} - 3 x^{2} y^{2} + 1} + \frac{- 2}{4 x^{3} - 3 x^{2} y^{2} + 1}$

Divide $x'$ by $1$ .

$x' = \frac{2 x^{3} y}{4 x^{3} - 3 x^{2} y^{2} + 1} + \frac{- 2}{4 x^{3} - 3 x^{2} y^{2} + 1}$

Simplify the right side.

Combine the numerators over the common denominator.

$x' = \frac{2 x^{3} y - 2}{4 x^{3} - 3 x^{2} y^{2} + 1}$

Factor $2$ out of $2 x^{3} y - 2$ .

Factor $2$ out of $2 x^{3} y$ .

$x' = \frac{2 (x^{3} y) - 2}{4 x^{3} - 3 x^{2} y^{2} + 1}$

Factor $2$ out of $- 2$ .

$x' = \frac{2 (x^{3} y) + 2 (- 1)}{4 x^{3} - 3 x^{2} y^{2} + 1}$

Factor $2$ out of $2 (x^{3} y) + 2 (- 1)$ .

$x' = \frac{2 (x^{3} y - 1)}{4 x^{3} - 3 x^{2} y^{2} + 1}$

Step 6

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

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

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

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