Calculus Examples

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

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 $e^{- y} (1 + \frac{d y}{d x}) = 1$ by $e^{- y}$ and simplify.

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Step 1.1.1.1

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

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

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.

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

Step 1.1.1.2.1.2

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

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

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

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Step 2.2.2.1

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

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

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Step 2.2.2.1.3.2.1

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

$0 - (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d y} [- y])$

Step 2.2.2.1.3.2.2

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

$0 - (e^{u_{1}} \frac{d}{d y} [- y])$

Step 2.2.2.1.3.2.3

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

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

Step 2.2.2.1.3.3

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

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

Step 2.2.2.1.3.4

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

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

Step 2.2.2.1.3.5

Multiply $- 1$ by $1$ .

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

Step 2.2.2.1.3.6

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

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

Step 2.2.2.1.3.7

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

$0 - - e^{- y}$

Step 2.2.2.1.3.8

Multiply $- 1$ by $- 1$ .

$0 + 1 e^{- y}$

Step 2.2.2.1.3.9

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

$0 + e^{- y}$

Step 2.2.2.1.4

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

$e^{- y}$

Step 2.2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Replace all occurrences of $u_{2}$ 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 $C$ .

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