Algebra Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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 $g (y) = y^{4}$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Move $4$ to the left of $x$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$4 x y^{3} + y^{4} x' - 1 \cdot 1$

Step 2.3.3

Multiply $- 1$ by $1$ .

$4 x y^{3} + y^{4} x' - 1$

Step 2.4

Reorder terms.

$4 y^{3} x + y^{4} x' - 1$

Step 3

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

$x'$

Step 4

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

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

Step 5

Solve for $x'$ .

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Step 5.1

Subtract $x'$ from both sides of the equation.

$4 y^{3} x + y^{4} x' - 1 - x' = 0$

Step 5.2

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

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Step 5.2.1

Subtract $4 y^{3} x$ from both sides of the equation.

$y^{4} x' - 1 - x' = - 4 y^{3} x$

Step 5.2.2

Add $1$ to both sides of the equation.

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

Rewrite $y^{4}$ as ${(y^{2})}^{2}$ .

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

Step 5.5

Rewrite $1$ as $1^{2}$ .

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

Step 5.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y^{2}$ and $b = 1$ .

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

Step 5.7

Factor.

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Step 5.7.1

Simplify.

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Step 5.7.1.1

Rewrite $1$ as $1^{2}$ .

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

Step 5.7.1.2

Factor.

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Step 5.7.1.2.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 1$ .

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

Step 5.7.1.2.2

Remove unnecessary parentheses.

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

Step 5.7.2

Remove unnecessary parentheses.

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

Step 5.8

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

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Step 5.8.1

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

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

Step 5.8.2

Simplify the left side.

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Step 5.8.2.1

Cancel the common factor of $y^{2} + 1$ .

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Step 5.8.2.1.1

Cancel the common factor.

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

Step 5.8.2.1.2

Rewrite the expression.

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

Step 5.8.2.2

Cancel the common factor of $y + 1$ .

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Step 5.8.2.2.1

Cancel the common factor.

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

Step 5.8.2.2.2

Rewrite the expression.

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

Step 5.8.2.3

Cancel the common factor of $y - 1$ .

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Step 5.8.2.3.1

Cancel the common factor.

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

Step 5.8.2.3.2

Divide $x'$ by $1$ .

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

Step 5.8.3

Simplify the right side.

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Step 5.8.3.1

Move the negative in front of the fraction.

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

Step 6

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

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