Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate using the Sum Rule.

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Step 2.1.1

Apply the product rule to $x y$ .

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $y$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

Replace all occurrences of $u$ with $y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$2 x^{2} y y' + 2 y^{2} x + 3 \cdot 1$

Step 2.3.3

Multiply $3$ by $1$ .

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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Step 3.1.1

To apply the Chain Rule, set $u$ as $y$ .

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

Step 3.1.2

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

$2 u \frac{d}{d x} [y]$

Step 3.1.3

Replace all occurrences of $u$ with $y$ .

$2 y \frac{d}{d x} [y]$

Step 3.2

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

$2 y y'$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $2 y y'$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

Subtract $3$ from both sides of the equation.

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

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

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

Step 5.5

Factor.

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Step 5.5.1

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

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

Step 5.5.2

Remove unnecessary parentheses.

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

Step 5.6

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

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Step 5.6.1

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

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

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

Cancel the common factor of $2$ .

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Step 5.6.2.1.1

Cancel the common factor.

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

Step 5.6.2.1.2

Rewrite the expression.

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

Step 5.6.2.2

Cancel the common factor of $y$ .

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Step 5.6.2.2.1

Cancel the common factor.

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

Step 5.6.2.2.2

Rewrite the expression.

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

Step 5.6.2.3

Cancel the common factor of $x + 1$ .

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Step 5.6.2.3.1

Cancel the common factor.

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

Step 5.6.2.3.2

Rewrite the expression.

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

Step 5.6.2.4

Cancel the common factor of $x - 1$ .

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Step 5.6.2.4.1

Cancel the common factor.

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

Step 5.6.2.4.2

Divide $y'$ by $1$ .

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

Step 5.6.3

Simplify the right side.

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Step 5.6.3.1

Simplify each term.

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Step 5.6.3.1.1

Cancel the common factor of $- 2$ and $2$ .

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Step 5.6.3.1.1.1

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

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

Step 5.6.3.1.1.2

Cancel the common factors.

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Step 5.6.3.1.1.2.1

Factor $2$ out of $2 y (x + 1) (x - 1)$ .

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

Step 5.6.3.1.1.2.2

Cancel the common factor.

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

Step 5.6.3.1.1.2.3

Rewrite the expression.

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

Step 5.6.3.1.2

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

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Step 5.6.3.1.2.1

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

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

Step 5.6.3.1.2.2

Cancel the common factors.

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Step 5.6.3.1.2.2.1

Factor $y$ out of $(y (x + 1)) (x - 1)$ .

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

Step 5.6.3.1.2.2.2

Cancel the common factor.

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

Step 5.6.3.1.2.2.3

Rewrite the expression.

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

Step 5.6.3.1.3

Move the negative in front of the fraction.

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

Step 5.6.3.1.4

Move the negative in front of the fraction.

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

Step 6

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

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