Calculus Examples

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

Step 1

Let $v = e^{x + y}$ . Substitute $v$ for all occurrences of $e^{x + y}$ .

$\frac{d x}{d y} + 1 = v$

Step 2

Find $\frac{d v}{d y}$ by differentiating $e^{x + y}$ .

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Step 2.1

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 $g (y) = x + y$ .

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Step 2.1.1

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

$\frac{d v}{d y} = \frac{d}{d u} [e^{u}] \frac{d}{d y} [x + y]$

Step 2.1.2

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

$\frac{d v}{d y} = e^{u} \frac{d}{d y} [x + y]$

Step 2.1.3

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

$\frac{d v}{d y} = e^{x + y} \frac{d}{d y} [x + y]$

Step 2.2

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

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

Step 2.3

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

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

Step 2.4

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

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

Step 3

Substitute $v$ for $e^{x + y}$ .

$\frac{d v}{d y} = v (x' + 1)$

Step 4

Substitute the derivative back in to the differential equation.

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Step 4.1

Add $- 1$ and $1$ .

$\frac{\frac{d v}{d y}}{v} + 0 = v$

Step 4.2

Add $\frac{\frac{d v}{d y}}{v}$ and $0$ .

$\frac{\frac{d v}{d y}}{v} = v$

Step 5

Separate the variables.

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Step 5.1

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

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Step 5.1.1

Multiply both sides by $v$ .

$\frac{\frac{d v}{d y}}{v} v = v \cdot v$

Step 5.1.2

Simplify.

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Step 5.1.2.1

Simplify the left side.

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Step 5.1.2.1.1

Cancel the common factor of $v$ .

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Step 5.1.2.1.1.1

Cancel the common factor.

$\frac{\frac{d v}{d y}}{v} v = v \cdot v$

Step 5.1.2.1.1.2

Rewrite the expression.

$\frac{d v}{d y} = v \cdot v$

Step 5.1.2.2

Simplify the right side.

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Step 5.1.2.2.1

Multiply $v$ by $v$ .

$\frac{d v}{d y} = v^{2}$

Step 5.2

Multiply both sides by $\frac{1}{v^{2}}$ .

$\frac{1}{v^{2}} \frac{d v}{d y} = \frac{1}{v^{2}} v^{2}$

Step 5.3

Cancel the common factor of $v^{2}$ .

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Step 5.3.1

Cancel the common factor.

$\frac{1}{v^{2}} \frac{d v}{d y} = \frac{1}{v^{2}} v^{2}$

Step 5.3.2

Rewrite the expression.

$\frac{1}{v^{2}} \frac{d v}{d y} = 1$

Step 5.4

Rewrite the equation.

$\frac{1}{v^{2}} d v = 1 d y$

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

$\int \frac{1}{v^{2}} d v = \int d y$

Step 6.2

Integrate the left side.

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Step 6.2.1

Apply basic rules of exponents.

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Step 6.2.1.1

Move $v^{2}$ out of the denominator by raising it to the $- 1$ power.

$\int {(v^{2})}^{- 1} d v = \int d y$

Step 6.2.1.2

Multiply the exponents in ${(v^{2})}^{- 1}$ .

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Step 6.2.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int v^{2 \cdot - 1} d v = \int d y$

Step 6.2.1.2.2

Multiply $2$ by $- 1$ .

$\int v^{- 2} d v = \int d y$

Step 6.2.2

By the Power Rule, the integral of $v^{- 2}$ with respect to $v$ is $- v^{- 1}$ .

$- v^{- 1} + C_{1} = \int d y$

Step 6.2.3

Rewrite $- v^{- 1} + C_{1}$ as $- \frac{1}{v} + C_{1}$ .

$- \frac{1}{v} + C_{1} = \int d y$

Step 6.3

Apply the constant rule.

$- \frac{1}{v} + C_{1} = y + C_{2}$

Step 6.4

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

$- \frac{1}{v} = y + K$

Step 7

Solve for $v$ .

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Step 7.1

Find the LCD of the terms in the equation.

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Step 7.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$v, 1, 1$

Step 7.1.2

The LCM of one and any expression is the expression.

$v$

Step 7.2

Multiply each term in $- \frac{1}{v} = y + K$ by $v$ to eliminate the fractions.

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Step 7.2.1

Multiply each term in $- \frac{1}{v} = y + K$ by $v$ .

$- \frac{1}{v} v = y v + K v$

Step 7.2.2

Simplify the left side.

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Step 7.2.2.1

Cancel the common factor of $v$ .

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Step 7.2.2.1.1

Move the leading negative in $- \frac{1}{v}$ into the numerator.

$\frac{- 1}{v} v = y v + K v$

Step 7.2.2.1.2

Cancel the common factor.

$\frac{- 1}{v} v = y v + K v$

Step 7.2.2.1.3

Rewrite the expression.

$- 1 = y v + K v$

Step 7.3

Solve the equation.

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Step 7.3.1

Rewrite the equation as $y v + K v = - 1$ .

$y v + K v = - 1$

Step 7.3.2

Factor $v$ out of $y v + K v$ .

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Step 7.3.2.1

Factor $v$ out of $y v$ .

$v y + K v = - 1$

Step 7.3.2.2

Factor $v$ out of $K v$ .

$v y + v K = - 1$

Step 7.3.2.3

Factor $v$ out of $v y + v K$ .

$v (y + K) = - 1$

Step 7.3.3

Divide each term in $v (y + K) = - 1$ by $y + K$ and simplify.

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Step 7.3.3.1

Divide each term in $v (y + K) = - 1$ by $y + K$ .

$\frac{v (y + K)}{y + K} = \frac{- 1}{y + K}$

Step 7.3.3.2

Simplify the left side.

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Step 7.3.3.2.1

Cancel the common factor of $y + K$ .

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Step 7.3.3.2.1.1

Cancel the common factor.

$\frac{v (y + K)}{y + K} = \frac{- 1}{y + K}$

Step 7.3.3.2.1.2

Divide $v$ by $1$ .

$v = \frac{- 1}{y + K}$

Step 7.3.3.3

Simplify the right side.

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Step 7.3.3.3.1

Move the negative in front of the fraction.

$v = - \frac{1}{y + K}$

Step 8

Replace all occurrences of $v$ with $e^{x + y}$ .

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

Step 9

Solve for $x$ .

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Step 9.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 9.2

Expand the left side.

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Step 9.2.1

Expand $\ln (e^{x + y})$ by moving $x + y$ outside the logarithm.

$(x + y) \ln (e) = \ln (- \frac{1}{y + K})$

Step 9.2.2

The natural logarithm of $e$ is $1$ .

$(x + y) \cdot 1 = \ln (- \frac{1}{y + K})$

Step 9.2.3

Multiply $x + y$ by $1$ .

$x + y = \ln (- \frac{1}{y + K})$

Step 9.3

Subtract $y$ from both sides of the equation.

$x = \ln (- \frac{1}{y + K}) - y$