Calculus Examples

$\frac{d y}{d x} = 1 + 6 x e^{x - y}$

Step 1

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

$\frac{d y}{d x} = 1 + 6 x v$

Step 2

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

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 3

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

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

Step 4

Substitute the derivative back in to the differential equation.

$- \frac{\frac{d v}{d x}}{v} + 1 = 1 + 6 x v$

Step 5

Separate the variables.

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Step 5.1

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

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Step 5.1.1

Move all terms not containing $\frac{d v}{d x}$ to the right side of the equation.

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Step 5.1.1.1

Subtract $1$ from both sides of the equation.

$- \frac{\frac{d v}{d x}}{v} = 1 + 6 x v - 1$

Step 5.1.1.2

Combine the opposite terms in $1 + 6 x v - 1$ .

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Step 5.1.1.2.1

Subtract $1$ from $1$ .

$- \frac{\frac{d v}{d x}}{v} = 6 x v + 0$

Step 5.1.1.2.2

Add $6 x v$ and $0$ .

$- \frac{\frac{d v}{d x}}{v} = 6 x v$

Step 5.1.2

Divide each term in $- \frac{\frac{d v}{d x}}{v} = 6 x v$ by $- 1$ and simplify.

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Step 5.1.2.1

Divide each term in $- \frac{\frac{d v}{d x}}{v} = 6 x v$ by $- 1$ .

$\frac{- \frac{\frac{d v}{d x}}{v}}{- 1} = \frac{6 x v}{- 1}$

Step 5.1.2.2

Simplify the left side.

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Step 5.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{\frac{\frac{d v}{d x}}{v}}{1} = \frac{6 x v}{- 1}$

Step 5.1.2.2.2

Divide $\frac{\frac{d v}{d x}}{v}$ by $1$ .

$\frac{\frac{d v}{d x}}{v} = \frac{6 x v}{- 1}$

Step 5.1.2.3

Simplify the right side.

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Step 5.1.2.3.1

Move the negative one from the denominator of $\frac{6 x v}{- 1}$ .

$\frac{\frac{d v}{d x}}{v} = - 1 \cdot (6 x v)$

Step 5.1.2.3.2

Rewrite $- 1 \cdot (6 x v)$ as $- (6 x v)$ .

$\frac{\frac{d v}{d x}}{v} = - (6 x v)$

Step 5.1.2.3.3

Multiply $6$ by $- 1$ .

$\frac{\frac{d v}{d x}}{v} = - 6 x v$

Step 5.1.3

Multiply both sides by $v$ .

$\frac{\frac{d v}{d x}}{v} v = - 6 x v \cdot v$

Step 5.1.4

Simplify.

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Step 5.1.4.1

Simplify the left side.

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Step 5.1.4.1.1

Cancel the common factor of $v$ .

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Step 5.1.4.1.1.1

Cancel the common factor.

$\frac{\frac{d v}{d x}}{v} v = - 6 x v \cdot v$

Step 5.1.4.1.1.2

Rewrite the expression.

$\frac{d v}{d x} = - 6 x v \cdot v$

Step 5.1.4.2

Simplify the right side.

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Step 5.1.4.2.1

Multiply $v$ by $v$ by adding the exponents.

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Step 5.1.4.2.1.1

Move $v$ .

$\frac{d v}{d x} = - 6 x (v \cdot v)$

Step 5.1.4.2.1.2

Multiply $v$ by $v$ .

$\frac{d v}{d x} = - 6 x v^{2}$

Step 5.2

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

$\frac{1}{v^{2}} \frac{d v}{d x} = \frac{1}{v^{2}} (- 6 x v^{2})$

Step 5.3

Simplify.

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Step 5.3.1

Rewrite using the commutative property of multiplication.

$\frac{1}{v^{2}} \frac{d v}{d x} = - 6 \frac{1}{v^{2}} (x v^{2})$

Step 5.3.2

Combine $- 6$ and $\frac{1}{v^{2}}$ .

$\frac{1}{v^{2}} \frac{d v}{d x} = \frac{- 6}{v^{2}} (x v^{2})$

Step 5.3.3

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

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Step 5.3.3.1

Factor $v^{2}$ out of $x v^{2}$ .

$\frac{1}{v^{2}} \frac{d v}{d x} = \frac{- 6}{v^{2}} (v^{2} x)$

Step 5.3.3.2

Cancel the common factor.

$\frac{1}{v^{2}} \frac{d v}{d x} = \frac{- 6}{v^{2}} (v^{2} x)$

Step 5.3.3.3

Rewrite the expression.

$\frac{1}{v^{2}} \frac{d v}{d x} = - 6 x$

Step 5.4

Rewrite the equation.

$\frac{1}{v^{2}} d v = - 6 x d x$

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

$\int \frac{1}{v^{2}} d v = \int - 6 x d x$

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.

$\int {(v^{2})}^{- 1} d v = \int - 6 x d x$

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, ${(a^{m})}^{n} = a^{m n}$ .

$\int v^{2 \cdot - 1} d v = \int - 6 x d x$

Step 6.2.1.2.2

Multiply $2$ by $- 1$ .

$\int v^{- 2} d v = \int - 6 x d x$

Step 6.2.2

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

$- v^{- 1} + C_{1} = \int - 6 x d x$

Step 6.2.3

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

$- \frac{1}{v} + C_{1} = \int - 6 x d x$

Step 6.3

Integrate the right side.

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Step 6.3.1

Since $- 6$ is constant with respect to $x$ , move $- 6$ out of the integral.

$- \frac{1}{v} + C_{1} = - 6 \int x d x$

Step 6.3.2

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$- \frac{1}{v} + C_{1} = - 6 (\frac{1}{2} x^{2} + C_{2})$

Step 6.3.3

Simplify the answer.

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Step 6.3.3.1

Rewrite $- 6 (\frac{1}{2} x^{2} + C_{2})$ as $- 6 (\frac{1}{2}) x^{2} + C_{2}$ .

$- \frac{1}{v} + C_{1} = - 6 (\frac{1}{2}) x^{2} + C_{2}$

Step 6.3.3.2

Simplify.

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Step 6.3.3.2.1

Combine $- 6$ and $\frac{1}{2}$ .

$- \frac{1}{v} + C_{1} = \frac{- 6}{2} x^{2} + C_{2}$

Step 6.3.3.2.2

Cancel the common factor of $- 6$ and $2$ .

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Step 6.3.3.2.2.1

Factor $2$ out of $- 6$ .

$- \frac{1}{v} + C_{1} = \frac{2 \cdot - 3}{2} x^{2} + C_{2}$

Step 6.3.3.2.2.2

Cancel the common factors.

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Step 6.3.3.2.2.2.1

Factor $2$ out of $2$ .

$- \frac{1}{v} + C_{1} = \frac{2 \cdot - 3}{2 (1)} x^{2} + C_{2}$

Step 6.3.3.2.2.2.2

Cancel the common factor.

$- \frac{1}{v} + C_{1} = \frac{2 \cdot - 3}{2 \cdot 1} x^{2} + C_{2}$

Step 6.3.3.2.2.2.3

Rewrite the expression.

$- \frac{1}{v} + C_{1} = \frac{- 3}{1} x^{2} + C_{2}$

Step 6.3.3.2.2.2.4

Divide $- 3$ by $1$ .

$- \frac{1}{v} + C_{1} = - 3 x^{2} + C_{2}$

Step 6.4

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

$- \frac{1}{v} = - 3 x^{2} + K$

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 $- \frac{1}{v} = - 3 x^{2} + K$ by $v$ to eliminate the fractions.

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Step 7.2.1

Multiply each term in $- \frac{1}{v} = - 3 x^{2} + K$ by $v$ .

$- \frac{1}{v} v = - 3 x^{2} v + K v$

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.

$\frac{- 1}{v} v = - 3 x^{2} v + K v$

Step 7.2.2.1.2

Cancel the common factor.

$\frac{- 1}{v} v = - 3 x^{2} v + K v$

Step 7.2.2.1.3

Rewrite the expression.

$- 1 = - 3 x^{2} v + K v$

Step 7.3

Solve the equation.

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Step 7.3.1

Rewrite the equation as $- 3 x^{2} v + K v = - 1$ .

$- 3 x^{2} v + K v = - 1$

Step 7.3.2

Factor $v$ out of $- 3 x^{2} v + K v$ .

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Step 7.3.2.1

Factor $v$ out of $- 3 x^{2} v$ .

$v (- 3 x^{2}) + K (v) = - 1$

Step 7.3.2.2

Factor $v$ out of $K v$ .

$v (- 3 x^{2}) + v (K) = - 1$

Step 7.3.2.3

Factor $v$ out of $v (- 3 x^{2}) + v (K)$ .

$v (- 3 x^{2} + K) = - 1$

Step 7.3.3

Divide each term in $v (- 3 x^{2} + K) = - 1$ by $- 3 x^{2} + K$ and simplify.

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Step 7.3.3.1

Divide each term in $v (- 3 x^{2} + K) = - 1$ by $- 3 x^{2} + K$ .

$\frac{v (- 3 x^{2} + K)}{- 3 x^{2} + K} = \frac{- 1}{- 3 x^{2} + K}$

Step 7.3.3.2

Simplify the left side.

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Step 7.3.3.2.1

Cancel the common factor of $- 3 x^{2} + K$ .

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Step 7.3.3.2.1.1

Cancel the common factor.

$\frac{v (- 3 x^{2} + K)}{- 3 x^{2} + K} = \frac{- 1}{- 3 x^{2} + K}$

Step 7.3.3.2.1.2

Divide $v$ by $1$ .

$v = \frac{- 1}{- 3 x^{2} + K}$

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.

$v = - \frac{1}{- 3 x^{2} + K}$

Step 7.3.3.3.2

Factor $- 1$ out of $- 3 x^{2}$ .

$v = - \frac{1}{- (3 x^{2}) + K}$

Step 7.3.3.3.3

Factor $- 1$ out of $K$ .

$v = - \frac{1}{- (3 x^{2}) - 1 (- K)}$

Step 7.3.3.3.4

Factor $- 1$ out of $- (3 x^{2}) - 1 (- K)$ .

$v = - \frac{1}{- (3 x^{2} - K)}$

Step 7.3.3.3.5

Simplify the expression.

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Step 7.3.3.3.5.1

Rewrite $- (3 x^{2} - K)$ as $- 1 (3 x^{2} - K)$ .

$v = - \frac{1}{- 1 (3 x^{2} - K)}$

Step 7.3.3.3.5.2

Move the negative in front of the fraction.

$v = - - \frac{1}{3 x^{2} - K}$

Step 7.3.3.3.5.3

Multiply $- 1$ by $- 1$ .

$v = 1 \frac{1}{3 x^{2} - K}$

Step 7.3.3.3.5.4

Multiply $\frac{1}{3 x^{2} - K}$ by $1$ .

$v = \frac{1}{3 x^{2} - K}$

Step 8

Simplify the constant of integration.

$v = \frac{1}{3 x^{2} + K}$

Step 9

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

$e^{x - y} = \frac{1}{3 x^{2} + K}$

Step 10

Solve for $y$ .

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Step 10.1

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

$\ln (e^{x - y}) = \ln (\frac{1}{3 x^{2} + K})$

Step 10.2

Expand the left side.

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Step 10.2.1

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

$(x - y) \ln (e) = \ln (\frac{1}{3 x^{2} + K})$

Step 10.2.2

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

$(x - y) \cdot 1 = \ln (\frac{1}{3 x^{2} + K})$

Step 10.2.3

Multiply $x - y$ by $1$ .

$x - y = \ln (\frac{1}{3 x^{2} + K})$

Step 10.3

Subtract $x$ from both sides of the equation.

$- y = \ln (\frac{1}{3 x^{2} + K}) - x$

Step 10.4

Divide each term in $- y = \ln (\frac{1}{3 x^{2} + K}) - x$ by $- 1$ and simplify.

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Step 10.4.1

Divide each term in $- y = \ln (\frac{1}{3 x^{2} + K}) - x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\ln (\frac{1}{3 x^{2} + K})}{- 1} + \frac{- x}{- 1}$

Step 10.4.2

Simplify the left side.

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Step 10.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\ln (\frac{1}{3 x^{2} + K})}{- 1} + \frac{- x}{- 1}$

Step 10.4.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (\frac{1}{3 x^{2} + K})}{- 1} + \frac{- x}{- 1}$

Step 10.4.3

Simplify the right side.

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Step 10.4.3.1

Simplify each term.

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Step 10.4.3.1.1

Move the negative one from the denominator of $\frac{\ln (\frac{1}{3 x^{2} + K})}{- 1}$ .

$y = - 1 \cdot \ln (\frac{1}{3 x^{2} + K}) + \frac{- x}{- 1}$

Step 10.4.3.1.2

Rewrite $- 1 \cdot \ln (\frac{1}{3 x^{2} + K})$ as $- \ln (\frac{1}{3 x^{2} + K})$ .

$y = - \ln (\frac{1}{3 x^{2} + K}) + \frac{- x}{- 1}$

Step 10.4.3.1.3

Dividing two negative values results in a positive value.

$y = - \ln (\frac{1}{3 x^{2} + K}) + \frac{x}{1}$

Step 10.4.3.1.4

Divide $x$ by $1$ .

$y = - \ln (\frac{1}{3 x^{2} + K}) + x$