Calculus Examples

$y (1 + \cos (x y)) d x + x (1 + \cos (x y)) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = y (1 + \cos (x y))$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y (1 + \cos (x y))]$

Step 1.2

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 $f (y) = y$ and $g (y) = 1 + \cos (x y)$ .

$\frac{\partial M}{\partial y} = y \frac{d}{d y} [1 + \cos (x y)] + (1 + \cos (x y)) \frac{d}{d y} [y]$

Step 1.3

Differentiate.

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = y (\frac{d}{d y} [1] + \frac{d}{d y} [\cos (x y)]) + (1 + \cos (x y)) \frac{d}{d y} [y]$

Step 1.3.2

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = y (0 + \frac{d}{d y} [\cos (x y)]) + (1 + \cos (x y)) \frac{d}{d y} [y]$

Step 1.3.3

Add $0$ and $\frac{d}{d y} [\cos (x y)]$ .

$\frac{\partial M}{\partial y} = y \frac{d}{d y} [\cos (x y)] + (1 + \cos (x y)) \frac{d}{d y} [y]$

Step 1.4

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 1.4.1

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

$\frac{\partial M}{\partial y} = y (\frac{d}{d u} [\cos (u)] \frac{d}{d y} [x y]) + (1 + \cos (x y)) \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = y (- \sin (u) \frac{d}{d y} [x y]) + (1 + \cos (x y)) \frac{d}{d y} [y]$

Step 1.4.3

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

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

Step 1.5

Differentiate.

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Step 1.5.1

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

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

Step 1.5.2

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

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

Step 1.5.3

Multiply $x$ by $1$ .

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

Step 1.5.4

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

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

Step 1.5.5

Simplify the expression.

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Step 1.5.5.1

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

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

Step 1.5.5.2

Reorder terms.

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x (1 + \cos (x y))$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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 $g (x) = 1 + \cos (x y)$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 2.3.3

Add $0$ and $\frac{d}{d x} [\cos (x y)]$ .

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

Step 2.4

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 2.4.1

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

$\frac{\partial N}{\partial x} = x (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [x y]) + (1 + \cos (x y)) \frac{d}{d x} [x]$

Step 2.4.2

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

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

Step 2.4.3

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

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

Step 2.5

Differentiate.

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Step 2.5.1

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

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

Step 2.5.2

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

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

Step 2.5.3

Multiply $y$ by $1$ .

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

Step 2.5.4

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

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

Step 2.5.5

Simplify the expression.

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Step 2.5.5.1

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

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

Step 2.5.5.2

Reorder terms.

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

Step 3

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

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Step 3.1

Substitute $- x y \sin (x y) + 1 + \cos (x y)$ for $\frac{\partial M}{\partial y}$ and $- x y \sin (x y) + 1 + \cos (x y)$ for $\frac{\partial N}{\partial x}$ .

$- x y \sin (x y) + 1 + \cos (x y) = - x y \sin (x y) + 1 + \cos (x y)$

Step 3.2

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

$- x y \sin (x y) + 1 + \cos (x y) = - x y \sin (x y) + 1 + \cos (x y)$ is an identity.

Step 4

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

$f (x, y) = \int x (1 + \cos (x y)) d y$

Step 5

Integrate $N (x, y) = x (1 + \cos (x y))$ to find $f (x, y)$ .

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Step 5.1

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

$f (x, y) = x \int 1 + \cos (x y) d y$

Step 5.2

Split the single integral into multiple integrals.

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

Step 5.3

Apply the constant rule.

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

Step 5.4

Let $u = x y$ . Then $d u = x d y$ , so $\frac{1}{x} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.4.1

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

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Step 5.4.1.1

Differentiate $x y$ .

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

Step 5.4.1.2

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

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

Step 5.4.1.3

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

$x \cdot 1$

Step 5.4.1.4

Multiply $x$ by $1$ .

$x$

Step 5.4.2

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

$f (x, y) = x (y + C + \int \cos (u) \frac{1}{x} d u)$

Step 5.5

Combine $\cos (u)$ and $\frac{1}{x}$ .

$f (x, y) = x (y + C + \int \frac{\cos (u)}{x} d u)$

Step 5.6

Since $\frac{1}{x}$ is constant with respect to $u$ , move $\frac{1}{x}$ out of the integral.

$f (x, y) = x (y + C + \frac{1}{x} \int \cos (u) d u)$

Step 5.7

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

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

Step 5.8

Simplify.

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

Step 5.9

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

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = y (1 + \cos (x y))$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [x (y + \frac{\sin (x y)}{x}) + g (x)] = y (1 + \cos (x y))$

Step 8.2

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

$\frac{d}{d x} [x (y + \frac{\sin (x y)}{x})] + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3

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

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Step 8.3.1

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 $f (x) = x$ and $g (x) = y + \frac{\sin (x y)}{x}$ .

$x \frac{d}{d x} [y + \frac{\sin (x y)}{x}] + (y + \frac{\sin (x y)}{x}) \frac{d}{d x} [x] + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3.2

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

$x (\frac{d}{d x} [y] + \frac{d}{d x} [\frac{\sin (x y)}{x}]) + (y + \frac{\sin (x y)}{x}) \frac{d}{d x} [x] + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3.3

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

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

Step 8.3.4

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = \sin (x y)$ and $g (x) = x$ .

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

Step 8.3.5

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

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Step 8.3.5.1

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

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

Step 8.3.5.2

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

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

Step 8.3.5.3

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

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

Step 8.3.6

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

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

Step 8.3.7

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

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

Step 8.3.8

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

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

Step 8.3.9

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

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

Step 8.3.10

Multiply $y$ by $1$ .

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

Step 8.3.11

Multiply $- 1$ by $1$ .

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

Step 8.3.12

Add $0$ and $\frac{x (\cos (x y) y) - \sin (x y)}{x^{2}}$ .

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

Step 8.3.13

Combine $x$ and $\frac{x (\cos (x y) y) - \sin (x y)}{x^{2}}$ .

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

Step 8.3.14

Cancel the common factors.

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Step 8.3.14.1

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

$\frac{x (x (\cos (x y) y) - \sin (x y))}{x \cdot x} + (y + \frac{\sin (x y)}{x}) \cdot 1 + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3.14.2

Cancel the common factor.

$\frac{x (x (\cos (x y) y) - \sin (x y))}{x \cdot x} + (y + \frac{\sin (x y)}{x}) \cdot 1 + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3.14.3

Rewrite the expression.

$\frac{x (\cos (x y) y) - \sin (x y)}{x} + (y + \frac{\sin (x y)}{x}) \cdot 1 + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3.15

Multiply $y + \frac{\sin (x y)}{x}$ by $1$ .

$\frac{x (\cos (x y) y) - \sin (x y)}{x} + y + \frac{\sin (x y)}{x} + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3.16

Combine the numerators over the common denominator.

$y + \frac{x (\cos (x y) y) - \sin (x y) + \sin (x y)}{x} + \frac{d}{d x} [g (x)] = y (1 + \cos (x y))$

Step 8.3.17

Add $- \sin (x y)$ and $\sin (x y)$ .

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

Step 8.3.18

Add $x (\cos (x y) y)$ and $0$ .

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

Step 8.3.19

Cancel the common factor of $x$ .

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Step 8.3.19.1

Cancel the common factor.

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

Step 8.3.19.2

Divide $\cos (x y) y$ by $1$ .

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

Step 8.4

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

$y + \cos (x y) y + \frac{d g}{d x} = y (1 + \cos (x y))$

Step 8.5

Reorder terms.

$\frac{d g}{d x} + y \cos (x y) + y = y (1 + \cos (x y))$

Step 9

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

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Step 9.1

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

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Step 9.1.1

Simplify the right side.

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Step 9.1.1.1

Simplify $y (1 + \cos (x y))$ .

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Step 9.1.1.1.1

Rewrite.

$\frac{d g}{d x} + y \cos (x y) + y = 0 + 0 + y (1 + \cos (x y))$

Step 9.1.1.1.2

Simplify by adding zeros.

$\frac{d g}{d x} + y \cos (x y) + y = y (1 + \cos (x y))$

Step 9.1.1.1.3

Apply the distributive property.

$\frac{d g}{d x} + y \cos (x y) + y = y \cdot 1 + y \cos (x y)$

Step 9.1.1.1.4

Multiply $y$ by $1$ .

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

Step 9.1.2

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

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Step 9.1.2.1

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

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

Step 9.1.2.2

Subtract $y$ from both sides of the equation.

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

Step 9.1.2.3

Combine the opposite terms in $y + y \cos (x y) - y \cos (x y) - y$ .

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Step 9.1.2.3.1

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

$\frac{d g}{d x} = y + 0 - y$

Step 9.1.2.3.2

Add $y$ and $0$ .

$\frac{d g}{d x} = y - y$

Step 9.1.2.3.3

Subtract $y$ from $y$ .

$\frac{d g}{d x} = 0$

Step 10

Find the antiderivative of $0$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 0$ .

$\int \frac{d g}{d x} d x = \int 0 d x$

Step 10.2

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

$g (x) = \int 0 d x$

Step 10.3

The integral of $0$ with respect to $x$ is $0$ .

$g (x) = 0 + C$

Step 10.4

Add $0$ and $C$ .

$g (x) = C$

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

Apply the distributive property.

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

Step 12.2

Cancel the common factor of $x$ .

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Step 12.2.1

Cancel the common factor.

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

Step 12.2.2

Rewrite the expression.

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