Calculus Examples

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

Step 1

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

Tap for more steps...

Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

Tap for more steps...

Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

Tap for more steps...

Step 1.3.1

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

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

Step 1.3.2

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

Tap for more steps...

Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.2.3

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

Multiply $- 1$ by $1$ .

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

Step 1.3.6

Move $- 1$ to the left of $e^{- y}$ .

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

Step 1.3.7

Rewrite $- 1 e^{- y}$ as $- e^{- y}$ .

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

Step 1.3.8

Multiply $- 1$ by $2$ .

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

Step 1.4

Simplify.

Tap for more steps...

Step 1.4.1

Subtract $2 x e^{- y}$ from $0$ .

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

Step 1.4.2

Reorder the factors of $- 2 x e^{- y}$ .

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

Step 1.4.3

Reorder factors in $- 2 e^{- y} x$ .

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

Step 2

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

Tap for more steps...

Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 2.4.1

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

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

Step 2.4.2

Add $e^{x}$ and $0$ .

$\frac{\partial N}{\partial x} = e^{x}$

Step 3

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

Tap for more steps...

Step 3.1

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

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

Step 3.2

Since the left side does not equal the right side, the equation is not an identity.

$- 2 x e^{- y} = e^{x}$ is not an identity.

Step 4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

Tap for more steps...

Step 4.1

Substitute $e^{x}$ for $\frac{\partial N}{\partial x}$ .

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

Step 4.2

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

$\frac{e^{x} - (- 2 x e^{- y})}{M}$

Step 4.3

Substitute $e^{x} + 2 x e^{- y}$ for $M$ .

Tap for more steps...

Step 4.3.1

Substitute $e^{x} + 2 x e^{- y}$ for $M$ .

$\frac{e^{x} - (- 2 x e^{- y})}{e^{x} + 2 x e^{- y}}$

Step 4.3.2

Multiply $- 2$ by $- 1$ .

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

Step 4.3.3

Substitute $e^{x} + 2 x e^{- y}$ for $M$ .

Tap for more steps...

Step 4.3.3.1

Cancel the common factor.

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

Step 4.3.3.2

Rewrite the expression.

$1$

Step 4.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

$μ (x, y) = e^{\int d y}$

Step 5

Evaluate the integral $e^{\int d y}$ .

Tap for more steps...

Step 5.1

Apply the constant rule.

$μ (x, y) = e^{y + C}$

Step 5.2

Simplify.

$μ (x, y) = e^{y} + C$

Step 6

Multiply both sides of $(e^{x} + 2 x e^{- y}) d x + (e^{x} + e^{- y}) d y = 0$ by the integration factor $e^{y}$ .

Tap for more steps...

Step 6.1

Multiply $e^{x} + 2 x e^{- y}$ by $e^{y}$ .

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

Step 6.2

Apply the distributive property.

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

Step 6.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.4

Multiply $e^{- y}$ by $e^{y}$ by adding the exponents.

Tap for more steps...

Step 6.4.1

Move $e^{y}$ .

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

Step 6.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.4.3

Subtract $y$ from $y$ .

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

Step 6.5

Simplify $2 x e^{0}$ .

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

Step 6.6

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

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

Step 6.7

Apply the distributive property.

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

Step 6.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.9

Multiply $e^{- y}$ by $e^{y}$ by adding the exponents.

Tap for more steps...

Step 6.9.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.9.2

Add $- y$ and $y$ .

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

Step 6.10

Simplify $e^{0}$ .

$(e^{x + y} + 2 x) d x + (e^{x + y} + 1) d y = 0$

Step 7

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

$f (x, y) = \int e^{x + y} + 1 d y$

Step 8

Integrate $N (x, y) = e^{x + y} + 1$ to find $f (x, y)$ .

Tap for more steps...

Step 8.1

Split the single integral into multiple integrals.

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

Step 8.2

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

Tap for more steps...

Step 8.2.1

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

Tap for more steps...

Step 8.2.1.1

Differentiate $x + y$ .

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

Step 8.2.1.2

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 8.2.1.3

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 8.2.1.4

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 8.2.1.5

Add $0$ and $1$ .

$1$

Step 8.2.2

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

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

Step 8.3

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

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

Step 8.4

Apply the constant rule.

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

Step 8.5

Simplify.

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

Step 8.6

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

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

Step 9

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

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

Step 10

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

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

Step 11

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

Tap for more steps...

Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

Tap for more steps...

Step 11.3.1

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 $g (x) = x + y$ .

Tap for more steps...

Step 11.3.1.1

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

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

Step 11.3.1.2

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

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

Step 11.3.1.3

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

Add $1$ and $0$ .

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

Step 11.3.6

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

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

Step 11.4

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

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

Step 11.5

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

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

Step 11.6

Simplify.

Tap for more steps...

Step 11.6.1

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

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

Step 11.6.2

Reorder terms.

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

Step 12

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

Tap for more steps...

Step 12.1

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

Tap for more steps...

Step 12.1.1

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

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

Step 12.1.2

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

Tap for more steps...

Step 12.1.2.1

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

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

Step 12.1.2.2

Add $2 x$ and $0$ .

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

Step 13

Find the antiderivative of $2 x$ to find $g (x)$ .

Tap for more steps...

Step 13.1

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

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

Step 13.2

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

$g (x) = \int 2 x d x$

Step 13.3

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

$g (x) = 2 \int x d x$

Step 13.4

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

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

Step 13.5

Simplify the answer.

Tap for more steps...

Step 13.5.1

Rewrite $2 (\frac{1}{2} x^{2} + C)$ as $2 (\frac{1}{2}) x^{2} + C$ .

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

Step 13.5.2

Simplify.

Tap for more steps...

Step 13.5.2.1

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

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

Step 13.5.2.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 13.5.2.2.1

Cancel the common factor.

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

Step 13.5.2.2.2

Rewrite the expression.

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

Step 13.5.2.3

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

$g (x) = x^{2} + C$

Step 14

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

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