Calculus Examples

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

Step 1

Add $(1 + e^{y}) d x$ to both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

Cancel the common factor.

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

Step 3.2.2

Rewrite the expression.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $1 + e^{y}$ .

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

Step 4.2.1.1.2

Differentiate.

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Step 4.2.1.1.2.1

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 4.2.1.1.2.2

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 4.2.1.1.3

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 4.2.1.1.4

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

$e^{y}$

Step 4.2.1.2

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

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

Step 4.2.2

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

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

Step 4.2.3

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

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

Step 4.3

Apply the constant rule.

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

Step 4.4

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

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