Calculus Examples

$x^{2} y + x y^{2} = 0$ , $(- 1, 1)$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

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

Step 2.2.3.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.3

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

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Step 2.3.1

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 $g (y) = y^{2}$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.4

Reorder terms.

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

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.

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

Step 5

Solve for $x'$ .

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

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

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

Factor $y x'$ out of $y x' (2 x) + y x' y$ .

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $y$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

Cancel the common factor of $2 x + y$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Divide $x'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

Move the negative in front of the fraction.

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

Step 5.3.3.1.2

Cancel the common factor of $y$ .

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Step 5.3.3.1.2.1

Cancel the common factor.

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

Step 5.3.3.1.2.2

Rewrite the expression.

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

Step 5.3.3.1.3

Move the negative in front of the fraction.

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

Step 5.3.3.2

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

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

Step 5.3.3.3

Write each expression with a common denominator of $y (2 x + y)$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.3.3.1

Multiply $\frac{2 x}{2 x + y}$ by $\frac{y}{y}$ .

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

Step 5.3.3.3.2

Reorder the factors of $(2 x + y) y$ .

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

Step 5.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.5

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

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Step 5.3.3.5.1

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

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

Step 5.3.3.5.2

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

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

Step 5.3.3.5.3

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

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

Step 5.3.3.6

Factor $- 1$ out of $- x$ .

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

Step 5.3.3.7

Factor $- 1$ out of $- 2 y$ .

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

Step 5.3.3.8

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

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

Step 5.3.3.9

Simplify the expression.

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Step 5.3.3.9.1

Rewrite $- (x + 2 y)$ as $- 1 (x + 2 y)$ .

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

Step 5.3.3.9.2

Move the negative in front of the fraction.

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

Step 6

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

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

Step 7

Replace $x$ with $- 1$ and $y$ with $1$ in the expression.

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

Step 8

Simplify the result.

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Step 8.1

Reduce the expression by cancelling the common factors.

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Step 8.1.1

Remove parentheses.

$- \frac{(- 1) ((- 1) + 2 (1))}{1 (2 (- 1) + 1)}$

Step 8.1.2

Cancel the common factor of $- 1$ and $1$ .

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Step 8.1.2.1

Rewrite $- 1$ as $- 1 (1)$ .

$- \frac{- 1 (1) ((- 1) + 2 (1))}{1 (2 (- 1) + 1)}$

Step 8.1.2.2

Cancel the common factor.

$- \frac{- 1 \cdot 1 (- 1 + 2 (1))}{1 (2 (- 1) + 1)}$

Step 8.1.2.3

Rewrite the expression.

$- \frac{- 1 (- 1 + 2 (1))}{2 (- 1) + 1}$

Step 8.2

Simplify the numerator.

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Step 8.2.1

Multiply $2$ by $1$ .

$- \frac{- 1 (- 1 + 2)}{2 (- 1) + 1}$

Step 8.2.2

Add $- 1$ and $2$ .

$- \frac{- 1 \cdot 1}{2 (- 1) + 1}$

Step 8.3

Simplify the denominator.

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Step 8.3.1

Multiply $2$ by $- 1$ .

$- \frac{- 1 \cdot 1}{- 2 + 1}$

Step 8.3.2

Add $- 2$ and $1$ .

$- \frac{- 1 \cdot 1}{- 1}$

Step 8.4

Multiply $- 1$ by $1$ .

$- \frac{- 1}{- 1}$

Step 8.5

Cancel the common factor.

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Step 8.5.1

Dividing two negative values results in a positive value.

$- \frac{1}{1}$

Step 8.5.2

Cancel the common factor of $1$ .

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Step 8.5.2.1

Cancel the common factor.

$- \frac{1}{1}$

Step 8.5.2.2

Rewrite the expression.

$- 1 \cdot 1$

Step 8.6

Multiply $- 1$ by $1$ .

$- 1$