Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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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 $g (x) = y$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

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

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

Step 2.4

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

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

Step 2.5

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

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Step 2.5.1

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 2.5.1.1

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

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

Step 2.5.1.2

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

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

Step 2.5.1.3

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

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

Step 2.5.2

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

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

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

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

Step 2.6.2

Multiply $2$ by $- 1$ .

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

Step 2.6.3

Reorder terms.

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

Step 3

Since $0$ is constant with respect to $x$ , the derivative of $0$ with respect to $x$ is $0$ .

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

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

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

Step 5.1.3

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

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.2.4

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

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

Step 5.2.5

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

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.

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

Step 5.3.3.1.2

Move the negative in front of the fraction.

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

Step 5.3.3.2

Combine into one fraction.

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Step 5.3.3.2.1

Combine the numerators over the common denominator.

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

Step 5.3.3.2.2

Combine the numerators over the common denominator.

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

Step 6

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

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