Calculus Examples

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

Step 1

Subtract $(1 + e^{2 y}) d x$ from both sides of the equation.

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

Step 2

Multiply both sides by $\frac{1}{x (1 + e^{2 y})}$ .

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

Combine $2$ and $\frac{1}{x (1 + e^{2 y})}$ .

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

Step 3.3

Cancel the common factor of $x$ .

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Step 3.3.1

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

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

Step 3.3.2

Cancel the common factor.

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

Step 3.3.3

Rewrite the expression.

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

Step 3.4

Combine $\frac{2}{1 + e^{2 y}}$ and $e^{2 y}$ .

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

Step 3.5

Rewrite using the commutative property of multiplication.

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

Step 3.6

Cancel the common factor of $1 + e^{2 y}$ .

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Step 3.6.1

Move the leading negative in $- \frac{1}{x (1 + e^{2 y})}$ into the numerator.

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

Step 3.6.2

Factor $1 + e^{2 y}$ out of $x (1 + e^{2 y})$ .

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

Step 3.6.3

Cancel the common factor.

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

Step 3.6.4

Rewrite the expression.

$\frac{2 e^{2 y}}{1 + e^{2 y}} d y = \frac{- 1}{x} d x$

Step 3.7

Move the negative in front of the fraction.

$\frac{2 e^{2 y}}{1 + e^{2 y}} d y = - \frac{1}{x} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{2 e^{2 y}}{1 + e^{2 y}} d y = \int - \frac{1}{x} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

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

$2 \int \frac{e^{2 y}}{1 + e^{2 y}} d y = \int - \frac{1}{x} d x$

Step 4.2.2

Let $u_{2} = 1 + e^{2 y}$ . Then $d u_{2} = 2 e^{2 y} d y$ , so $\frac{1}{2} d u_{2} = e^{2 y} d y$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.2.2.1

Let $u_{2} = 1 + e^{2 y}$ . Find $\frac{d u_{2}}{d y}$ .

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Step 4.2.2.1.1

Differentiate $1 + e^{2 y}$ .

$\frac{d}{d y} [1 + e^{2 y}]$

Step 4.2.2.1.2

Differentiate.

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Step 4.2.2.1.2.1

By the Sum Rule, the derivative of $1 + e^{2 y}$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [e^{2 y}]$ .

$\frac{d}{d y} [1] + \frac{d}{d y} [e^{2 y}]$

Step 4.2.2.1.2.2

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

$0 + \frac{d}{d y} [e^{2 y}]$

Step 4.2.2.1.3

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

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Step 4.2.2.1.3.1

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

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Step 4.2.2.1.3.1.1

To apply the Chain Rule, set $u_{1}$ as $2 y$ .

$0 + \frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d y} [2 y]$

Step 4.2.2.1.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$0 + e^{u_{1}} \frac{d}{d y} [2 y]$

Step 4.2.2.1.3.1.3

Replace all occurrences of $u_{1}$ with $2 y$ .

$0 + e^{2 y} \frac{d}{d y} [2 y]$

Step 4.2.2.1.3.2

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

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

Step 4.2.2.1.3.3

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

$0 + e^{2 y} (2 \cdot 1)$

Step 4.2.2.1.3.4

Multiply $2$ by $1$ .

$0 + e^{2 y} \cdot 2$

Step 4.2.2.1.3.5

Move $2$ to the left of $e^{2 y}$ .

$0 + 2 e^{2 y}$

Step 4.2.2.1.4

Add $0$ and $2 e^{2 y}$ .

$2 e^{2 y}$

Step 4.2.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$2 \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2} = \int - \frac{1}{x} d x$

Step 4.2.3

Simplify.

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Step 4.2.3.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{2}$ .

$2 \int \frac{1}{u_{2} \cdot 2} d u_{2} = \int - \frac{1}{x} d x$

Step 4.2.3.2

Move $2$ to the left of $u_{2}$ .

$2 \int \frac{1}{2 u_{2}} d u_{2} = \int - \frac{1}{x} d x$

Step 4.2.4

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

$2 (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2}) = \int - \frac{1}{x} d x$

Step 4.2.5

Simplify.

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Step 4.2.5.1

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

$\frac{2}{2} \int \frac{1}{u_{2}} d u_{2} = \int - \frac{1}{x} d x$

Step 4.2.5.2

Cancel the common factor of $2$ .

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Step 4.2.5.2.1

Cancel the common factor.

$\frac{2}{2} \int \frac{1}{u_{2}} d u_{2} = \int - \frac{1}{x} d x$

Step 4.2.5.2.2

Rewrite the expression.

$1 \int \frac{1}{u_{2}} d u_{2} = \int - \frac{1}{x} d x$

Step 4.2.5.3

Multiply $\int \frac{1}{u_{2}} d u_{2}$ by $1$ .

$\int \frac{1}{u_{2}} d u_{2} = \int - \frac{1}{x} d x$

Step 4.2.6

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\ln (| u_{2} |) + C_{1} = \int - \frac{1}{x} d x$

Step 4.2.7

Replace all occurrences of $u_{2}$ with $1 + e^{2 y}$ .

$\ln (| 1 + e^{2 y} |) + C_{1} = \int - \frac{1}{x} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$\ln (| 1 + e^{2 y} |) + C_{1} = - \int \frac{1}{x} d x$

Step 4.3.2

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$\ln (| 1 + e^{2 y} |) + C_{1} = - (\ln (| x |) + C_{2})$

Step 4.3.3

Simplify.

$\ln (| 1 + e^{2 y} |) + C_{1} = - \ln (| x |) + C_{2}$

Step 4.4

Group the constant of integration on the right side as $C$ .

$\ln (| 1 + e^{2 y} |) = - \ln (| x |) + C$