Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

Divide each term in $e^{y} (\frac{d y}{d x} + 1) = 1$ by $e^{y}$ and simplify.

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Step 1.1.1.1

Divide each term in $e^{y} (\frac{d y}{d x} + 1) = 1$ by $e^{y}$ .

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

Step 1.1.1.2

Simplify the left side.

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Step 1.1.1.2.1

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

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Step 1.1.1.2.1.1

Cancel the common factor.

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

Step 1.1.1.2.1.2

Divide $\frac{d y}{d x} + 1$ by $1$ .

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

Step 1.1.2

Subtract $1$ from both sides of the equation.

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

Step 1.2

Multiply both sides by $\frac{1}{\frac{1}{e^{y}} - 1}$ .

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

Step 1.3

Cancel the common factor of $\frac{1}{e^{y}} - 1$ .

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Step 1.3.1

Cancel the common factor.

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

Step 1.3.2

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify.

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Step 2.2.1.1

Simplify the denominator.

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Step 2.2.1.1.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{e^{y}}{e^{y}}$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

Combine the numerators over the common denominator.

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

Step 2.2.1.2

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.2.1.3

Multiply $\frac{e^{y}}{1 - e^{y}}$ by $1$ .

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

Step 2.2.2

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

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Step 2.2.2.1

Let $u = 1 - e^{y}$ . Find $\frac{d u}{d y}$ .

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Step 2.2.2.1.1

Differentiate $1 - e^{y}$ .

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

Step 2.2.2.1.2

Differentiate.

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Step 2.2.2.1.2.1

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

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

Step 2.2.2.1.2.2

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

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

Step 2.2.2.1.3

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

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Step 2.2.2.1.3.1

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

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

Step 2.2.2.1.3.2

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

$0 - e^{y}$

Step 2.2.2.1.4

Subtract $e^{y}$ from $0$ .

$- e^{y}$

Step 2.2.2.2

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

$\int \frac{- 1}{u} d u = \int d x$

Step 2.2.3

Split the fraction into multiple fractions.

$\int - \frac{1}{u} d u = \int d x$

Step 2.2.4

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

$- \int \frac{1}{u} d u = \int d x$

Step 2.2.5

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

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

Step 2.2.6

Simplify.

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

Step 2.2.7

Replace all occurrences of $u$ with $1 - e^{y}$ .

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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