Calculus Examples

$\cos (x + y) d x + (3 y^{2} + 2 y + \cos (x + y)) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = \cos (x + y)$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [\cos (x + y)]$

Step 1.2

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

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d u} [\cos (u)] \frac{d}{d y} [x + y]$

Step 1.2.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

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

Step 1.2.3

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

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Add $0$ and $\frac{d}{d y} [y]$ .

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

Step 1.3.4

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

$\frac{\partial M}{\partial y} = - \sin (x + y) \cdot 1$

Step 1.3.5

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = - \sin (x + y)$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 3 y^{2} + 2 y + \cos (x + y)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [2 y] + \frac{d}{d x} [\cos (x + y)]$

Step 2.2.3

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

$\frac{\partial N}{\partial x} = 0 + 0 + \frac{d}{d x} [\cos (x + y)]$

Step 2.3

Evaluate $\frac{d}{d x} [\cos (x + y)]$ .

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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 $f (x) = \cos (x)$ and $g (x) = x + y$ .

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Step 2.3.1.1

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

$\frac{\partial N}{\partial x} = 0 + 0 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [x + y]$

Step 2.3.1.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$\frac{\partial N}{\partial x} = 0 + 0 - \sin (u) \frac{d}{d x} [x + y]$

Step 2.3.1.3

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

$\frac{\partial N}{\partial x} = 0 + 0 - \sin (x + y) \frac{d}{d x} [x + y]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 + 0 - \sin (x + y) (\frac{d}{d x} [x] + \frac{d}{d x} [y])$

Step 2.3.3

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

$\frac{\partial N}{\partial x} = 0 + 0 - \sin (x + y) (1 + \frac{d}{d x} [y])$

Step 2.3.4

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

$\frac{\partial N}{\partial x} = 0 + 0 - \sin (x + y) (1 + 0)$

Step 2.3.5

Add $1$ and $0$ .

$\frac{\partial N}{\partial x} = 0 + 0 - \sin (x + y) \cdot 1$

Step 2.3.6

Multiply $- 1$ by $1$ .

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

Step 2.4

Combine terms.

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Step 2.4.1

Add $0$ and $0$ .

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

Step 2.4.2

Subtract $\sin (x + y)$ from $0$ .

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

Step 3

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

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Step 3.1

Substitute $- \sin (x + y)$ for $\frac{\partial M}{\partial y}$ and $- \sin (x + y)$ for $\frac{\partial N}{\partial x}$ .

$- \sin (x + y) = - \sin (x + y)$

Step 3.2

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

$- \sin (x + y) = - \sin (x + y)$ is an identity.

Step 4

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

$f (x, y) = \int \cos (x + y) d x$

Step 5

Integrate $M (x, y) = \cos (x + y)$ to find $f (x, y)$ .

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Step 5.1

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

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Step 5.1.1

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

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Step 5.1.1.1

Differentiate $x + y$ .

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

Step 5.1.1.2

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 5.1.1.3

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 5.1.1.4

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

$1 + 0$

Step 5.1.1.5

Add $1$ and $0$ .

$1$

Step 5.1.2

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

$f (x, y) = \int \cos (u) d u$

Step 5.2

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$f (x, y) = \sin (u) + C$

Step 5.3

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

$f (x, y) = \sin (x + y) + C$

Step 6

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

$f (x, y) = \sin (x + y) + g (y)$

Step 7

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

$\frac{\partial f}{\partial y} = 3 y^{2} + 2 y + \cos (x + y)$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

Evaluate $\frac{d}{d y} [\sin (x + y)]$ .

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Step 8.3.1

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

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Step 8.3.1.1

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

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

Step 8.3.1.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

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

Step 8.3.1.3

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

Add $0$ and $1$ .

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

Step 8.3.6

Multiply $\cos (x + y)$ by $1$ .

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

Step 8.4

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

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

Step 8.5

Reorder terms.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Subtract $\cos (x + y)$ from both sides of the equation.

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

Step 9.1.2

Combine the opposite terms in $3 y^{2} + 2 y + \cos (x + y) - \cos (x + y)$ .

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Step 9.1.2.1

Subtract $\cos (x + y)$ from $\cos (x + y)$ .

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

Step 9.1.2.2

Add $3 y^{2} + 2 y$ and $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

Split the single integral into multiple integrals.

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

Step 10.4

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

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

Step 10.5

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

$g (y) = 3 (\frac{1}{3} y^{3} + C) + \int 2 y d y$

Step 10.6

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

$g (y) = 3 (\frac{1}{3} y^{3} + C) + 2 \int y d y$

Step 10.7

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

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

Step 10.8

Simplify.

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

Step 10.9

Simplify.

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Step 10.9.1

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

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

Step 10.9.2

Cancel the common factor of $3$ .

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Step 10.9.2.1

Cancel the common factor.

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

Step 10.9.2.2

Rewrite the expression.

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

Step 10.9.3

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

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

Step 10.9.4

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

$g (y) = y^{3} + \frac{2}{2} y^{2} + C$

Step 10.9.5

Cancel the common factor of $2$ .

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Step 10.9.5.1

Cancel the common factor.

$g (y) = y^{3} + \frac{2}{2} y^{2} + C$

Step 10.9.5.2

Rewrite the expression.

$g (y) = y^{3} + 1 y^{2} + C$

Step 10.9.6

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

$g (y) = y^{3} + y^{2} + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = \sin (x + y) + g (y)$ .

$f (x, y) = \sin (x + y) + y^{3} + y^{2} + C$