Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

Evaluate $\frac{d}{d y} [\cos (2 y)]$ .

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Step 1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \cos (y)$ and $g (y) = 2 y$ .

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Step 1.3.1.1

To apply the Chain Rule, set $u$ as $2 y$ .

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

Step 1.3.1.2

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

$\frac{\partial M}{\partial y} = - \sin (u) \frac{d}{d y} [2 y] + \frac{d}{d y} [- 3 x^{2} y^{2}]$

Step 1.3.1.3

Replace all occurrences of $u$ with $2 y$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = - \sin (2 y) \cdot 2 + \frac{d}{d y} [- 3 x^{2} y^{2}]$

Step 1.3.5

Multiply $2$ by $- 1$ .

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

Step 1.4

Evaluate $\frac{d}{d y} [- 3 x^{2} y^{2}]$ .

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Step 1.4.1

Since $- 3 x^{2}$ is constant with respect to $y$ , the derivative of $- 3 x^{2} y^{2}$ with respect to $y$ is $- 3 x^{2} \frac{d}{d y} [y^{2}]$ .

$\frac{\partial M}{\partial y} = - 2 \sin (2 y) - 3 x^{2} \frac{d}{d y} [y^{2}]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = - 2 \sin (2 y) - 3 x^{2} (2 y)$

Step 1.4.3

Multiply $2$ by $- 3$ .

$\frac{\partial M}{\partial y} = - 2 \sin (2 y) - 6 x^{2} y$

Step 1.5

Reorder terms.

$\frac{\partial M}{\partial y} = - 6 x^{2} y - 2 \sin (2 y)$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [- 2 x \sin (2 y)] + \frac{d}{d x} [- 2 x^{3} y]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 0 - 2 \sin (2 y) \frac{d}{d x} [x] + \frac{d}{d x} [- 2 x^{3} y]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 - 2 \sin (2 y) \cdot 1 + \frac{d}{d x} [- 2 x^{3} y]$

Step 2.3.3

Multiply $- 2$ by $1$ .

$\frac{\partial N}{\partial x} = 0 - 2 \sin (2 y) + \frac{d}{d x} [- 2 x^{3} y]$

Step 2.4

Evaluate $\frac{d}{d x} [- 2 x^{3} y]$ .

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Step 2.4.1

Since $- 2 y$ is constant with respect to $x$ , the derivative of $- 2 x^{3} y$ with respect to $x$ is $- 2 y \frac{d}{d x} [x^{3}]$ .

$\frac{\partial N}{\partial x} = 0 - 2 \sin (2 y) - 2 y \frac{d}{d x} [x^{3}]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 0 - 2 \sin (2 y) - 2 y (3 x^{2})$

Step 2.4.3

Multiply $3$ by $- 2$ .

$\frac{\partial N}{\partial x} = 0 - 2 \sin (2 y) - 6 y x^{2}$

Step 2.5

Simplify.

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Step 2.5.1

Subtract $2 \sin (2 y)$ from $0$ .

$\frac{\partial N}{\partial x} = - 2 \sin (2 y) - 6 y x^{2}$

Step 2.5.2

Reorder terms.

$\frac{\partial N}{\partial x} = - 6 x^{2} y - 2 \sin (2 y)$

Step 3

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

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Step 3.1

Substitute $- 6 x^{2} y - 2 \sin (2 y)$ for $\frac{\partial M}{\partial y}$ and $- 6 x^{2} y - 2 \sin (2 y)$ for $\frac{\partial N}{\partial x}$ .

$- 6 x^{2} y - 2 \sin (2 y) = - 6 x^{2} y - 2 \sin (2 y)$

Step 3.2

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

$- 6 x^{2} y - 2 \sin (2 y) = - 6 x^{2} y - 2 \sin (2 y)$ is an identity.

Step 4

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

$f (x, y) = \int \cos (2 y) - 3 x^{2} y^{2} d x$

Step 5

Integrate $M (x, y) = \cos (2 y) - 3 x^{2} y^{2}$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int \cos (2 y) d x + \int - 3 x^{2} y^{2} d x$

Step 5.2

Apply the constant rule.

$f (x, y) = \cos (2 y) x + C + \int - 3 x^{2} y^{2} d x$

Step 5.3

Since $- 3 y^{2}$ is constant with respect to $x$ , move $- 3 y^{2}$ out of the integral.

$f (x, y) = \cos (2 y) x + C - 3 y^{2} \int x^{2} d x$

Step 5.4

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

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

Step 5.5

Simplify.

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

Step 5.6

Simplify.

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Step 5.6.1

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

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

Step 5.6.2

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

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Step 5.6.2.1

Factor $3$ out of $- 3$ .

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

Step 5.6.2.2

Cancel the common factors.

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Step 5.6.2.2.1

Factor $3$ out of $3$ .

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

Step 5.6.2.2.2

Cancel the common factor.

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

Step 5.6.2.2.3

Rewrite the expression.

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

Step 5.6.2.2.4

Divide $- 1$ by $1$ .

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

Evaluate $\frac{d}{d y} [\cos (2 y) x]$ .

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Step 8.3.1

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

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

Step 8.3.2

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 $g (y) = 2 y$ .

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Step 8.3.2.1

To apply the Chain Rule, set $u$ as $2 y$ .

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

Step 8.3.2.2

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

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

Step 8.3.2.3

Replace all occurrences of $u$ with $2 y$ .

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

Multiply $2$ by $1$ .

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

Step 8.3.6

Multiply $2$ by $- 1$ .

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

Step 8.4

Evaluate $\frac{d}{d y} [- y^{2} x^{3}]$ .

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Step 8.4.1

Since $- x^{3}$ is constant with respect to $y$ , the derivative of $- y^{2} x^{3}$ with respect to $y$ is $- x^{3} \frac{d}{d y} [y^{2}]$ .

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $2$ by $- 1$ .

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

Step 8.5

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

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

Step 8.6

Reorder terms.

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

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

Add $2 x \sin (2 y)$ to both sides of the equation.

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

Step 9.1.2

Add $2 x^{3} y$ to both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

Add $- 2 x \sin (2 y)$ and $2 x \sin (2 y)$ .

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

Step 9.1.3.2

Add $\cos (2 y) - 2 x^{3} y$ and $0$ .

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

Step 9.1.3.3

Add $- 2 x^{3} y$ and $2 x^{3} y$ .

$\frac{d g}{d y} = \cos (2 y) + 0$

Step 9.1.3.4

Add $\cos (2 y)$ and $0$ .

$\frac{d g}{d y} = \cos (2 y)$

Step 10

Find the antiderivative of $\cos (2 y)$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = \cos (2 y)$ .

$\int \frac{d g}{d y} d y = \int \cos (2 y) d y$

Step 10.2

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

$g (y) = \int \cos (2 y) d y$

Step 10.3

Let $u = 2 y$ . Then $d u = 2 d y$ , so $\frac{1}{2} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 10.3.1

Let $u = 2 y$ . Find $\frac{d u}{d y}$ .

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Step 10.3.1.1

Differentiate $2 y$ .

$\frac{d}{d y} [2 y]$

Step 10.3.1.2

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

$2 \frac{d}{d y} [y]$

Step 10.3.1.3

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

$2 \cdot 1$

Step 10.3.1.4

Multiply $2$ by $1$ .

$2$

Step 10.3.2

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

$g (y) = \int \cos (u) \frac{1}{2} d u$

Step 10.4

Combine $\cos (u)$ and $\frac{1}{2}$ .

$g (y) = \int \frac{\cos (u)}{2} d u$

Step 10.5

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$g (y) = \frac{1}{2} \int \cos (u) d u$

Step 10.6

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

$g (y) = \frac{1}{2} (\sin (u) + C)$

Step 10.7

Simplify.

$g (y) = \frac{1}{2} \sin (u) + C$

Step 10.8

Replace all occurrences of $u$ with $2 y$ .

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

Step 11

Substitute for $g (y)$ in $f (x, y) = \cos (2 y) x - y^{2} x^{3} + g (y)$ .

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

Step 12

Simplify $f (x, y) = \cos (2 y) x - y^{2} x^{3} + \frac{1}{2} \sin (2 y) + C$ .

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Step 12.1

Combine $\frac{1}{2}$ and $\sin (2 y)$ .

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

Step 12.2

Reorder factors in $f (x, y) = \cos (2 y) x - y^{2} x^{3} + \frac{\sin (2 y)}{2} + C$ .

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