Calculus Examples

$(3 x^{2} - 2 y + e^{x + y}) d x + (e^{x + y} - 2 x + 4) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 3 x^{2} - 2 y + e^{x + y}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [3 x^{2} - 2 y + e^{x + y}]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [3 x^{2}] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [e^{x + y}]$

Step 1.2.2

Since $3 x^{2}$ is constant with respect to $y$ , the derivative of $3 x^{2}$ with respect to $y$ is $0$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $- 2$ by $1$ .

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

Step 1.4

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

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Step 1.4.1

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

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Step 1.4.1.1

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

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

Step 1.4.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 1.4.1.3

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

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

Step 1.4.2

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

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

Step 1.4.3

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

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

Step 1.4.4

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

$\frac{\partial M}{\partial y} = 0 - 2 + e^{x + y} (0 + 1)$

Step 1.4.5

Add $0$ and $1$ .

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

Step 1.4.6

Multiply $e^{x + y}$ by $1$ .

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

Step 1.5

Subtract $2$ from $0$ .

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = e^{x + y} - 2 x + 4$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [e^{x + y} - 2 x + 4]$

Step 2.2

By the Sum Rule, the derivative of $e^{x + y} - 2 x + 4$ with respect to $x$ is $\frac{d}{d x} [e^{x + y}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [e^{x + y}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

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

$\frac{\partial N}{\partial x} = \frac{d}{d u} [e^{u}] \frac{d}{d x} [x + y] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{\partial N}{\partial x} = e^{u} \frac{d}{d x} [x + y] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3.1.3

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

$\frac{\partial N}{\partial x} = e^{x + y} \frac{d}{d x} [x + y] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = e^{x + y} (\frac{d}{d x} [x] + \frac{d}{d x} [y]) + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3.3

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

$\frac{\partial N}{\partial x} = e^{x + y} (1 + \frac{d}{d x} [y]) + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3.4

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

$\frac{\partial N}{\partial x} = e^{x + y} (1 + 0) + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3.5

Add $1$ and $0$ .

$\frac{\partial N}{\partial x} = e^{x + y} \cdot 1 + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.3.6

Multiply $e^{x + y}$ by $1$ .

$\frac{\partial N}{\partial x} = e^{x + y} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [4]$

Step 2.4

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

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = e^{x + y} - 2 \frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = e^{x + y} - 2 \cdot 1 + \frac{d}{d x} [4]$

Step 2.4.3

Multiply $- 2$ by $1$ .

$\frac{\partial N}{\partial x} = e^{x + y} - 2 + \frac{d}{d x} [4]$

Step 2.5

Differentiate using the Constant Rule.

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Step 2.5.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = e^{x + y} - 2 + 0$

Step 2.5.2

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

$- 2 + e^{x + y} = e^{x + y} - 2$ is an identity.

Step 4

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

$f (x, y) = \int e^{x + y} - 2 x + 4 d y$

Step 5

Integrate $N (x, y) = e^{x + y} - 2 x + 4$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int e^{x + y} d y + \int - 2 x d y + \int 4 d y$

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 $d u$ .

$f (x, y) = \int e^{u} d u + \int - 2 x d y + \int 4 d y$

Step 5.3

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$f (x, y) = e^{u} + C + \int - 2 x d y + \int 4 d y$

Step 5.4

Apply the constant rule.

$f (x, y) = e^{u} + C - 2 x y + C + \int 4 d y$

Step 5.5

Apply the constant rule.

$f (x, y) = e^{u} + C - 2 x y + C + 4 y + C$

Step 5.6

Simplify.

$f (x, y) = e^{u} - 2 x y + 4 y + C$

Step 5.7

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

$f (x, y) = e^{x + y} - 2 x y + 4 y + C$

Step 6

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

$f (x, y) = e^{x + y} - 2 x y + 4 y + g (x)$

Step 7

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

$\frac{\partial f}{\partial x} = 3 x^{2} - 2 y + e^{x + y}$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [e^{x + y} - 2 x y + 4 y + g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.2

By the Sum Rule, the derivative of $e^{x + y} - 2 x y + 4 y + g (x)$ with respect to $x$ is $\frac{d}{d x} [e^{x + y}] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [e^{x + y}] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.3

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

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Step 8.3.1

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

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Step 8.3.1.1

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

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [x + y] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$e^{u} \frac{d}{d x} [x + y] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.3.1.3

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

$e^{x + y} \frac{d}{d x} [x + y] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.3.2

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

$e^{x + y} (\frac{d}{d x} [x] + \frac{d}{d x} [y]) + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.3.3

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

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

Step 8.3.4

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

$e^{x + y} (1 + 0) + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.3.5

Add $1$ and $0$ .

$e^{x + y} \cdot 1 + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.3.6

Multiply $e^{x + y}$ by $1$ .

$e^{x + y} + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.4

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

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Step 8.4.1

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

$e^{x + y} - 2 y \frac{d}{d x} [x] + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.4.2

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

$e^{x + y} - 2 y \cdot 1 + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.4.3

Multiply $- 2$ by $1$ .

$e^{x + y} - 2 y + \frac{d}{d x} [4 y] + \frac{d}{d x} [g (x)] = 3 x^{2} - 2 y + e^{x + y}$

Step 8.5

Since $4 y$ is constant with respect to $x$ , the derivative of $4 y$ with respect to $x$ is $0$ .

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

Step 8.6

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

$e^{x + y} - 2 y + 0 + \frac{d g}{d x} = 3 x^{2} - 2 y + e^{x + y}$

Step 8.7

Simplify.

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Step 8.7.1

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

$e^{x + y} - 2 y + \frac{d g}{d x} = 3 x^{2} - 2 y + e^{x + y}$

Step 8.7.2

Reorder terms.

$\frac{d g}{d x} - 2 y + e^{x + y} = 3 x^{2} - 2 y + e^{x + y}$

Step 9

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

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Step 9.1

Move all terms not containing $\frac{d g}{d x}$ to the right side of the equation.

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Step 9.1.1

Add $2 y$ to both sides of the equation.

$\frac{d g}{d x} + e^{x + y} = 3 x^{2} - 2 y + e^{x + y} + 2 y$

Step 9.1.2

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

$\frac{d g}{d x} = 3 x^{2} - 2 y + e^{x + y} + 2 y - e^{x + y}$

Step 9.1.3

Combine the opposite terms in $3 x^{2} - 2 y + e^{x + y} + 2 y - e^{x + y}$ .

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Step 9.1.3.1

Add $- 2 y$ and $2 y$ .

$\frac{d g}{d x} = 3 x^{2} + 0 + e^{x + y} - e^{x + y}$

Step 9.1.3.2

Add $3 x^{2}$ and $0$ .

$\frac{d g}{d x} = 3 x^{2} + e^{x + y} - e^{x + y}$

Step 9.1.3.3

Subtract $e^{x + y}$ from $e^{x + y}$ .

$\frac{d g}{d x} = 3 x^{2} + 0$

Step 9.1.3.4

Add $3 x^{2}$ and $0$ .

$\frac{d g}{d x} = 3 x^{2}$

Step 10

Find the antiderivative of $3 x^{2}$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 3 x^{2}$ .

$\int \frac{d g}{d x} d x = \int 3 x^{2} d x$

Step 10.2

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

$g (x) = \int 3 x^{2} d x$

Step 10.3

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$g (x) = 3 \int x^{2} d x$

Step 10.4

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

$g (x) = 3 (\frac{1}{3} x^{3} + C)$

Step 10.5

Simplify the answer.

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Step 10.5.1

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

$g (x) = 3 (\frac{1}{3}) x^{3} + C$

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

$g (x) = \frac{3}{3} x^{3} + C$

Step 10.5.2.2

Cancel the common factor of $3$ .

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Step 10.5.2.2.1

Cancel the common factor.

$g (x) = \frac{3}{3} x^{3} + C$

Step 10.5.2.2.2

Rewrite the expression.

$g (x) = 1 x^{3} + C$

Step 10.5.2.3

Multiply $x^{3}$ by $1$ .

$g (x) = x^{3} + C$

Step 11

Substitute for $g (x)$ in $f (x, y) = e^{x + y} - 2 x y + 4 y + g (x)$ .

$f (x, y) = e^{x + y} - 2 x y + 4 y + x^{3} + C$