Find dx/dy x^4y^2-x^5y+5xy^3=0

$x^{4} y^{2} - x^{5} y + 5 x y^{3} = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (x^{4} y^{2} - x^{5} y + 5 x y^{3}) = \frac{d}{d y} (0)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $x^{4} y^{2} - x^{5} y + 5 x y^{3}$ with respect to $y$ is $\frac{d}{d y} [x^{4} y^{2}] + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$ .

$\frac{d}{d y} [x^{4} y^{2}] + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

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

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 $f (y) = x^{4}$ and $g (y) = y^{2}$ .

$x^{4} \frac{d}{d y} [y^{2}] + y^{2} \frac{d}{d y} [x^{4}] + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

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

$x^{4} (2 y) + y^{2} \frac{d}{d y} [x^{4}] + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

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

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

$x^{4} (2 y) + y^{2} (\frac{d}{d u_{1}} [{u_{1}}^{4}] \frac{d}{d y} [x]) + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

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

$x^{4} (2 y) + y^{2} (4 {u_{1}}^{3} \frac{d}{d y} [x]) + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

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

$x^{4} (2 y) + y^{2} (4 x^{3} \frac{d}{d y} [x]) + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

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

$x^{4} (2 y) + y^{2} (4 x^{3} x') + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

Move $2$ to the left of $x^{4}$ .

$2 \cdot x^{4} y + y^{2} (4 x^{3} x') + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

Move $4$ to the left of $y^{2}$ .

$2 x^{4} y + 4 y^{2} x^{3} x' + \frac{d}{d y} [- x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

Evaluate $\frac{d}{d y} [- x^{5} y]$ .

Since $- 1$ is constant with respect to $y$ , the derivative of $- x^{5} y$ with respect to $y$ is $- \frac{d}{d y} [x^{5} y]$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - \frac{d}{d y} [x^{5} y] + \frac{d}{d y} [5 x y^{3}]$

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 $f (y) = x^{5}$ and $g (y) = y$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} \frac{d}{d y} [y] + y \frac{d}{d y} [x^{5}]) + \frac{d}{d y} [5 x y^{3}]$

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

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} \cdot 1 + y \frac{d}{d y} [x^{5}]) + \frac{d}{d y} [5 x y^{3}]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{5}$ and $g (y) = x$ .

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

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} \cdot 1 + y (\frac{d}{d u_{2}} [{u_{2}}^{5}] \frac{d}{d y} [x])) + \frac{d}{d y} [5 x y^{3}]$

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 5$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} \cdot 1 + y (5 {u_{2}}^{4} \frac{d}{d y} [x])) + \frac{d}{d y} [5 x y^{3}]$

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

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} \cdot 1 + y (5 x^{4} \frac{d}{d y} [x])) + \frac{d}{d y} [5 x y^{3}]$

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

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} \cdot 1 + y (5 x^{4} x')) + \frac{d}{d y} [5 x y^{3}]$

Multiply $x^{5}$ by $1$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} + y (5 x^{4} x')) + \frac{d}{d y} [5 x y^{3}]$

Move $5$ to the left of $y$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} + 5 y x^{4} x') + \frac{d}{d y} [5 x y^{3}]$

Evaluate $\frac{d}{d y} [5 x y^{3}]$ .

Since $5$ is constant with respect to $y$ , the derivative of $5 x y^{3}$ with respect to $y$ is $5 \frac{d}{d y} [x y^{3}]$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} + 5 y x^{4} x') + 5 \frac{d}{d y} [x y^{3}]$

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

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} + 5 y x^{4} x') + 5 (x \frac{d}{d y} [y^{3}] + y^{3} \frac{d}{d y} [x])$

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

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} + 5 y x^{4} x') + 5 (x (3 y^{2}) + y^{3} \frac{d}{d y} [x])$

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

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} + 5 y x^{4} x') + 5 (x (3 y^{2}) + y^{3} x')$

Move $3$ to the left of $x$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - (x^{5} + 5 y x^{4} x') + 5 (3 x y^{2} + y^{3} x')$

Simplify.

Apply the distributive property.

$2 x^{4} y + 4 y^{2} x^{3} x' - x^{5} - (5 y x^{4} x') + 5 (3 x y^{2} + y^{3} x')$

Apply the distributive property.

$2 x^{4} y + 4 y^{2} x^{3} x' - x^{5} - (5 y x^{4} x') + 5 (3 x y^{2}) + 5 (y^{3} x')$

Combine terms.

Multiply $5$ by $- 1$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - x^{5} - 5 (y x^{4} x') + 5 (3 x y^{2}) + 5 (y^{3} x')$

Multiply $3$ by $5$ .

$2 x^{4} y + 4 y^{2} x^{3} x' - x^{5} - 5 y x^{4} x' + 15 x y^{2} + 5 y^{3} x'$

Reorder terms.

$- x^{5} + 2 x^{4} y + 4 x^{3} y^{2} x' - 5 x^{4} y x' + 15 y^{2} x + 5 y^{3} x'$

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.

$- x^{5} + 2 x^{4} y + 4 x^{3} y^{2} x' - 5 x^{4} y x' + 15 y^{2} x + 5 y^{3} x' = 0$

Step 5

Solve for $x'$ .

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

Add $x^{5}$ to both sides of the equation.

$2 x^{4} y + 4 x^{3} y^{2} x' - 5 x^{4} y x' + 15 y^{2} x + 5 y^{3} x' = x^{5}$

Subtract $2 x^{4} y$ from both sides of the equation.

$4 x^{3} y^{2} x' - 5 x^{4} y x' + 15 y^{2} x + 5 y^{3} x' = x^{5} - 2 x^{4} y$

Subtract $15 y^{2} x$ from both sides of the equation.

$4 x^{3} y^{2} x' - 5 x^{4} y x' + 5 y^{3} x' = x^{5} - 2 x^{4} y - 15 y^{2} x$

Factor $y x'$ out of $4 x^{3} y^{2} x' - 5 x^{4} y x' + 5 y^{3} x'$ .

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

$y x' (4 x^{3} y) - 5 x^{4} y x' + 5 y^{3} x' = x^{5} - 2 x^{4} y - 15 y^{2} x$

Factor $y x'$ out of $- 5 x^{4} y x'$ .

$y x' (4 x^{3} y) + y x' (- 5 x^{4}) + 5 y^{3} x' = x^{5} - 2 x^{4} y - 15 y^{2} x$

Factor $y x'$ out of $5 y^{3} x'$ .

$y x' (4 x^{3} y) + y x' (- 5 x^{4}) + y x' (5 y^{2}) = x^{5} - 2 x^{4} y - 15 y^{2} x$

Factor $y x'$ out of $y x' (4 x^{3} y) + y x' (- 5 x^{4})$ .

$y x' (4 x^{3} y - 5 x^{4}) + y x' (5 y^{2}) = x^{5} - 2 x^{4} y - 15 y^{2} x$

Factor $y x'$ out of $y x' (4 x^{3} y - 5 x^{4}) + y x' (5 y^{2})$ .

$y x' (4 x^{3} y - 5 x^{4} + 5 y^{2}) = x^{5} - 2 x^{4} y - 15 y^{2} x$

Divide each term in $y x' (4 x^{3} y - 5 x^{4} + 5 y^{2}) = x^{5} - 2 x^{4} y - 15 y^{2} x$ by $y (4 x^{3} y - 5 x^{4} + 5 y^{2})$ and simplify.

Divide each term in $y x' (4 x^{3} y - 5 x^{4} + 5 y^{2}) = x^{5} - 2 x^{4} y - 15 y^{2} x$ by $y (4 x^{3} y - 5 x^{4} + 5 y^{2})$ .

$\frac{y x' (4 x^{3} y - 5 x^{4} + 5 y^{2})}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 2 x^{4} y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 15 y^{2} x}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Simplify the left side.

Cancel the common factor of $y$ .

Cancel the common factor.

Rewrite the expression.

$\frac{x' (4 x^{3} y - 5 x^{4} + 5 y^{2})}{4 x^{3} y - 5 x^{4} + 5 y^{2}} = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 2 x^{4} y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 15 y^{2} x}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Cancel the common factor of $4 x^{3} y - 5 x^{4} + 5 y^{2}$ .

Cancel the common factor.

Divide $x'$ by $1$ .

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 2 x^{4} y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 15 y^{2} x}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $y$ .

Cancel the common factor.

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 2 x^{4} y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 15 y^{2} x}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Rewrite the expression.

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} + \frac{- 2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}} + \frac{- 15 y^{2} x}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Move the negative in front of the fraction.

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{(2) x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}} + \frac{- 15 y^{2} x}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

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

Factor $y$ out of $- 15 y^{2} x$ .

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}} + \frac{y (- 15 y x)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Cancel the common factors.

Cancel the common factor.

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}} + \frac{y (- 15 y x)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Rewrite the expression.

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}} + \frac{- 15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

Move the negative in front of the fraction.

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

To write $- \frac{2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}} \cdot \frac{y}{y} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

Write each expression with a common denominator of $y (4 x^{3} y - 5 x^{4} + 5 y^{2})$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2 x^{4}}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$ by $\frac{y}{y}$ .

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{2 x^{4} y}{(4 x^{3} y - 5 x^{4} + 5 y^{2}) y} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

Reorder the factors of $(4 x^{3} y - 5 x^{4} + 5 y^{2}) y$ .

$x' = \frac{x^{5}}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{2 x^{4} y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

Combine the numerators over the common denominator.

$x' = \frac{x^{5} - 2 x^{4} y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

Factor $x^{4}$ out of $x^{5} - 2 x^{4} y$ .

Factor $x^{4}$ out of $x^{5}$ .

$x' = \frac{x^{4} x - 2 x^{4} y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

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

$x' = \frac{x^{4} x + x^{4} (- 2 y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

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

$x' = \frac{x^{4} (x - 2 y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$

To write $- \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$x' = \frac{x^{4} (x - 2 y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}} \cdot \frac{y}{y}$

Write each expression with a common denominator of $y (4 x^{3} y - 5 x^{4} + 5 y^{2})$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{15 y x}{4 x^{3} y - 5 x^{4} + 5 y^{2}}$ by $\frac{y}{y}$ .

$x' = \frac{x^{4} (x - 2 y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x y}{(4 x^{3} y - 5 x^{4} + 5 y^{2}) y}$

Reorder the factors of $(4 x^{3} y - 5 x^{4} + 5 y^{2}) y$ .

$x' = \frac{x^{4} (x - 2 y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})} - \frac{15 y x y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Combine the numerators over the common denominator.

$x' = \frac{x^{4} (x - 2 y) - 15 y x y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Simplify the numerator.

Factor $x$ out of $x^{4} (x - 2 y) - 15 y x y$ .

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

$x' = \frac{x (x^{3} (x - 2 y)) - 15 y x y}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Factor $x$ out of $- 15 y x y$ .

$x' = \frac{x (x^{3} (x - 2 y)) + x (- 15 y \cdot y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Factor $x$ out of $x (x^{3} (x - 2 y)) + x (- 15 y \cdot y)$ .

$x' = \frac{x (x^{3} (x - 2 y) - 15 y \cdot y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Apply the distributive property.

$x' = \frac{x (x^{3} x + x^{3} (- 2 y) - 15 y \cdot y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Multiply $x^{3}$ by $x$ by adding the exponents.

Multiply $x^{3}$ by $x$ .

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

$x' = \frac{x (x^{3} x^{1} + x^{3} (- 2 y) - 15 y \cdot y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x' = \frac{x (x^{3 + 1} + x^{3} (- 2 y) - 15 y \cdot y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Add $3$ and $1$ .

$x' = \frac{x (x^{4} + x^{3} (- 2 y) - 15 y \cdot y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Rewrite using the commutative property of multiplication.

$x' = \frac{x (x^{4} - 2 x^{3} y - 15 y \cdot y)}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$x' = \frac{x (x^{4} - 2 x^{3} y - 15 (y \cdot y))}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Multiply $y$ by $y$ .

$x' = \frac{x (x^{4} - 2 x^{3} y - 15 y^{2})}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Step 6

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

$\frac{d x}{d y} = \frac{x (x^{4} - 2 x^{3} y - 15 y^{2})}{y (4 x^{3} y - 5 x^{4} + 5 y^{2})}$

Find dx/dy x^4y^2-x^5y+5xy^3=0

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