Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = a^{2} - 2 x y - y^{2}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [a^{2} - 2 x y - y^{2}]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [a^{2}] + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [- y^{2}]$

Step 1.2.2

Since $a^{2}$ is constant with respect to $y$ , the derivative of $a^{2}$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [- y^{2}]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 0 - 2 x \frac{d}{d y} [y] + \frac{d}{d y} [- y^{2}]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 - 2 x \cdot 1 + \frac{d}{d y} [- y^{2}]$

Step 1.3.3

Multiply $- 2$ by $1$ .

$\frac{\partial M}{\partial y} = 0 - 2 x + \frac{d}{d y} [- y^{2}]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 0 - 2 x - \frac{d}{d y} [y^{2}]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 0 - 2 x - (2 y)$

Step 1.4.3

Multiply $2$ by $- 1$ .

$\frac{\partial M}{\partial y} = 0 - 2 x - 2 y$

Step 1.5

Subtract $2 x$ from $0$ .

$\frac{\partial M}{\partial y} = - 2 x - 2 y$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = - {(x + y)}^{2}$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Rewrite ${(x + y)}^{2}$ as $(x + y) (x + y)$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- ((x + y) (x + y))]$

Step 2.3

Expand $(x + y) (x + y)$ using the FOIL Method.

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Step 2.3.1

Apply the distributive property.

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (x (x + y) + y (x + y))]$

Step 2.3.2

Apply the distributive property.

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (x \cdot x + x y + y (x + y))]$

Step 2.3.3

Apply the distributive property.

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (x \cdot x + x y + y x + y \cdot y)]$

Step 2.4

Simplify and combine like terms.

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Step 2.4.1

Simplify each term.

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Step 2.4.1.1

Multiply $x$ by $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (x^{2} + x y + y x + y \cdot y)]$

Step 2.4.1.2

Multiply $y$ by $y$ .

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

Step 2.4.2

Add $x y$ and $y x$ .

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Step 2.4.2.1

Reorder $y$ and $x$ .

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

Step 2.4.2.2

Add $x y$ and $x y$ .

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

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

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

Step 2.10

Multiply $2$ by $1$ .

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

Step 2.11

Since $y^{2}$ is constant with respect to $x$ , the derivative of $y^{2}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = - (2 x + 2 y + 0)$

Step 2.12

Add $2 x + 2 y$ and $0$ .

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

Step 2.13

Simplify.

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Step 2.13.1

Apply the distributive property.

$\frac{\partial N}{\partial x} = - (2 x) - (2 y)$

Step 2.13.2

Combine terms.

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Step 2.13.2.1

Multiply $2$ by $- 1$ .

$\frac{\partial N}{\partial x} = - 2 x - (2 y)$

Step 2.13.2.2

Multiply $2$ by $- 1$ .

$\frac{\partial N}{\partial x} = - 2 x - 2 y$

Step 3

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 3.1

Substitute $- 2 x - 2 y$ for $\frac{\partial M}{\partial y}$ and $- 2 x - 2 y$ for $\frac{\partial N}{\partial x}$ .

$- 2 x - 2 y = - 2 x - 2 y$

Step 3.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$- 2 x - 2 y = - 2 x - 2 y$ is an identity.

Step 4

Set $f (x, y)$ equal to the integral of $N (x, y)$ .

$f (x, y) = \int - {(x + y)}^{2} d y$

Step 5

Integrate $N (x, y) = - {(x + y)}^{2}$ to find $f (x, y)$ .

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Step 5.1

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$f (x, y) = - \int {(x + y)}^{2} d y$

Step 5.2

Let $u = x + y$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.2.1

Let $u = x + y$ . Find $\frac{d u}{d y}$ .

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Step 5.2.1.1

Differentiate $x + y$ .

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

Step 5.2.1.2

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

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

Step 5.2.1.3

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

$0 + \frac{d}{d y} [y]$

Step 5.2.1.4

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

$0 + 1$

Step 5.2.1.5

Add $0$ and $1$ .

$1$

Step 5.2.2

Rewrite the problem using $u$ and $d u$ .

$f (x, y) = - \int u^{2} d u$

Step 5.3

By the Power Rule, the integral of $u^{2}$ with respect to $u$ is $\frac{1}{3} u^{3}$ .

$f (x, y) = - (\frac{1}{3} u^{3} + C)$

Step 5.4

Rewrite $- (\frac{1}{3} u^{3} + C)$ as $- \frac{1}{3} u^{3} + C$ .

$f (x, y) = - \frac{1}{3} u^{3} + C$

Step 5.5

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

$f (x, y) = - \frac{1}{3} {(x + y)}^{3} + C$

Step 6

Since the integral of $g (x)$ will contain an integration constant, we can replace $C$ with $g (x)$ .

$f (x, y) = - \frac{1}{3} {(x + y)}^{3} + g (x)$

Step 7

Set $\frac{\partial f}{\partial x} = M (x, y)$ .

$\frac{\partial f}{\partial x} = a^{2} - 2 x y - y^{2}$

Step 8

Find $\frac{\partial f}{\partial x}$ .

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [- \frac{1}{3} {(x + y)}^{3} + g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.2

By the Sum Rule, the derivative of $- \frac{1}{3} {(x + y)}^{3} + g (x)$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{3} {(x + y)}^{3}] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [- \frac{1}{3} {(x + y)}^{3}] + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3

Evaluate $\frac{d}{d x} [- \frac{1}{3} {(x + y)}^{3}]$ .

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Step 8.3.1

Since $- \frac{1}{3}$ is constant with respect to $x$ , the derivative of $- \frac{1}{3} {(x + y)}^{3}$ with respect to $x$ is $- \frac{1}{3} \frac{d}{d x} [{(x + y)}^{3}]$ .

$- \frac{1}{3} \frac{d}{d x} [{(x + y)}^{3}] + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = x + y$ .

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Step 8.3.2.1

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

$- \frac{1}{3} (\frac{d}{d u} [u^{3}] \frac{d}{d x} [x + y]) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.2.2

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

$- \frac{1}{3} (3 u^{2} \frac{d}{d x} [x + y]) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.2.3

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

$- \frac{1}{3} (3 {(x + y)}^{2} \frac{d}{d x} [x + y]) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.3

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

$- \frac{1}{3} (3 {(x + y)}^{2} (\frac{d}{d x} [x] + \frac{d}{d x} [y])) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.4

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

$- \frac{1}{3} (3 {(x + y)}^{2} (1 + \frac{d}{d x} [y])) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.5

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

$- \frac{1}{3} (3 {(x + y)}^{2} (1 + 0)) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.6

Add $1$ and $0$ .

$- \frac{1}{3} (3 {(x + y)}^{2} \cdot 1) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.7

Multiply $3$ by $1$ .

$- \frac{1}{3} (3 {(x + y)}^{2}) + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.8

Multiply $3$ by $- 1$ .

$- 3 (\frac{1}{3}) {(x + y)}^{2} + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.9

Combine $- 3$ and $\frac{1}{3}$ .

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

Step 8.3.10

Combine $\frac{- 3}{3}$ and ${(x + y)}^{2}$ .

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

Step 8.3.11

Cancel the common factor of $- 3$ and $3$ .

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Step 8.3.11.1

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

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

Step 8.3.11.2

Cancel the common factors.

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Step 8.3.11.2.1

Factor $3$ out of $3$ .

$\frac{3 (- {(x + y)}^{2})}{3 (1)} + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.11.2.2

Cancel the common factor.

$\frac{3 (- {(x + y)}^{2})}{3 \cdot 1} + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.11.2.3

Rewrite the expression.

$\frac{- {(x + y)}^{2}}{1} + \frac{d}{d x} [g (x)] = a^{2} - 2 x y - y^{2}$

Step 8.3.11.2.4

Divide $- {(x + y)}^{2}$ by $1$ .

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

Step 8.4

Differentiate using the function rule which states that the derivative of $g (x)$ is $\frac{d g}{d x}$ .

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

Step 8.5

Reorder terms.

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

Step 9

Solve for $\frac{d g}{d x}$ .

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Step 9.1

Add ${(x + y)}^{2}$ to both sides of the equation.

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

Step 10

Find the antiderivative of $a^{2} - 2 x y - y^{2} + {(x + y)}^{2}$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = a^{2} - 2 x y - y^{2} + {(x + y)}^{2}$ .

$\int \frac{d g}{d x} d x = \int a^{2} - 2 x y - y^{2} + {(x + y)}^{2} d x$

Step 10.2

Evaluate $\int \frac{d g}{d x} d x$ .

$g (x) = \int a^{2} - 2 x y - y^{2} + {(x + y)}^{2} d x$

Step 10.3

Split the single integral into multiple integrals.

$g (x) = \int a^{2} d x + \int - 2 x y d x + \int - y^{2} d x + \int {(x + y)}^{2} d x$

Step 10.4

Apply the constant rule.

$g (x) = a^{2} x + C + \int - 2 x y d x + \int - y^{2} d x + \int {(x + y)}^{2} d x$

Step 10.5

Since $- 2 y$ is constant with respect to $x$ , move $- 2 y$ out of the integral.

$g (x) = a^{2} x + C - 2 y \int x d x + \int - y^{2} d x + \int {(x + y)}^{2} d x$

Step 10.6

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$g (x) = a^{2} x + C - 2 y (\frac{1}{2} x^{2} + C) + \int - y^{2} d x + \int {(x + y)}^{2} d x$

Step 10.7

Apply the constant rule.

$g (x) = a^{2} x + C - 2 y (\frac{1}{2} x^{2} + C) - y^{2} x + C + \int {(x + y)}^{2} d x$

Step 10.8

Combine $\frac{1}{2}$ and $x^{2}$ .

$g (x) = a^{2} x + C - 2 y (\frac{x^{2}}{2} + C) - y^{2} x + C + \int {(x + y)}^{2} d x$

Step 10.9

Let $u = x + y$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 10.9.1

Let $u = x + y$ . Find $\frac{d u}{d x}$ .

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Step 10.9.1.1

Differentiate $x + y$ .

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

Step 10.9.1.2

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

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

Step 10.9.1.3

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

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

Step 10.9.1.4

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

$1 + 0$

Step 10.9.1.5

Add $1$ and $0$ .

$1$

Step 10.9.2

Rewrite the problem using $u$ and $d u$ .

$g (x) = a^{2} x + C - 2 y (\frac{x^{2}}{2} + C) - y^{2} x + C + \int u^{2} d u$

Step 10.10

By the Power Rule, the integral of $u^{2}$ with respect to $u$ is $\frac{1}{3} u^{3}$ .

$g (x) = a^{2} x + C - 2 y (\frac{x^{2}}{2} + C) - y^{2} x + C + \frac{1}{3} u^{3} + C$

Step 10.11

Simplify.

$g (x) = a^{2} x - y x^{2} - y^{2} x + \frac{1}{3} u^{3} + C$

Step 10.12

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

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

Step 11

Substitute for $g (x)$ in $f (x, y) = - \frac{1}{3} {(x + y)}^{3} + g (x)$ .

$f (x, y) = - \frac{1}{3} {(x + y)}^{3} + a^{2} x - y x^{2} - y^{2} x + \frac{1}{3} {(x + y)}^{3} + C$

Step 12

Combine the opposite terms in $- \frac{1}{3} {(x + y)}^{3} + a^{2} x - y x^{2} - y^{2} x + \frac{1}{3} {(x + y)}^{3} + C$ .

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Step 12.1

Add $- \frac{1}{3} {(x + y)}^{3}$ and $\frac{1}{3} {(x + y)}^{3}$ .

$f (x, y) = a^{2} x - y x^{2} - y^{2} x + 0 + C$

Step 12.2

Add $a^{2} x - y x^{2} - y^{2} x$ and $0$ .

$f (x, y) = a^{2} x - y x^{2} - y^{2} x + C$