Calculus Examples

$\frac{d y}{d x} = \frac{4}{x^{2} e^{3 y + 9}}$

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

$\frac{d y}{d x} = \frac{4}{x^{2}} \cdot \frac{1}{e^{3 y + 9}}$

Step 1.2

Multiply both sides by $e^{3 y + 9}$ .

$e^{3 y + 9} \frac{d y}{d x} = e^{3 y + 9} (\frac{4}{x^{2}} \cdot \frac{1}{e^{3 y + 9}})$

Step 1.3

Simplify.

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Step 1.3.1

Combine.

$e^{3 y + 9} \frac{d y}{d x} = e^{3 y + 9} \frac{4 \cdot 1}{x^{2} e^{3 y + 9}}$

Step 1.3.2

Cancel the common factor of $e^{3 y + 9}$ .

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Step 1.3.2.1

Factor $e^{3 y + 9}$ out of $x^{2} e^{3 y + 9}$ .

$e^{3 y + 9} \frac{d y}{d x} = e^{3 y + 9} \frac{4 \cdot 1}{e^{3 y + 9} x^{2}}$

Step 1.3.2.2

Cancel the common factor.

$e^{3 y + 9} \frac{d y}{d x} = e^{3 y + 9} \frac{4 \cdot 1}{e^{3 y + 9} x^{2}}$

Step 1.3.2.3

Rewrite the expression.

$e^{3 y + 9} \frac{d y}{d x} = \frac{4 \cdot 1}{x^{2}}$

Step 1.3.3

Multiply $4$ by $1$ .

$e^{3 y + 9} \frac{d y}{d x} = \frac{4}{x^{2}}$

Step 1.4

Rewrite the equation.

$e^{3 y + 9} d y = \frac{4}{x^{2}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int e^{3 y + 9} d y = \int \frac{4}{x^{2}} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 3 y + 9$ . Then $d u = 3 d y$ , so $\frac{1}{3} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 3 y + 9$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $3 y + 9$ .

$\frac{d}{d y} [3 y + 9]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $3 y + 9$ with respect to $y$ is $\frac{d}{d y} [3 y] + \frac{d}{d y} [9]$ .

$\frac{d}{d y} [3 y] + \frac{d}{d y} [9]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d y} [3 y]$ .

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Step 2.2.1.1.3.1

Since $3$ is constant with respect to $y$ , the derivative of $3 y$ with respect to $y$ is $3 \frac{d}{d y} [y]$ .

$3 \frac{d}{d y} [y] + \frac{d}{d y} [9]$

Step 2.2.1.1.3.2

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

$3 \cdot 1 + \frac{d}{d y} [9]$

Step 2.2.1.1.3.3

Multiply $3$ by $1$ .

$3 + \frac{d}{d y} [9]$

Step 2.2.1.1.4

Differentiate using the Constant Rule.

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Step 2.2.1.1.4.1

Since $9$ is constant with respect to $y$ , the derivative of $9$ with respect to $y$ is $0$ .

$3 + 0$

Step 2.2.1.1.4.2

Add $3$ and $0$ .

$3$

Step 2.2.1.2

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

$\int e^{u} \frac{1}{3} d u = \int \frac{4}{x^{2}} d x$

Step 2.2.2

Combine $e^{u}$ and $\frac{1}{3}$ .

$\int \frac{e^{u}}{3} d u = \int \frac{4}{x^{2}} d x$

Step 2.2.3

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int e^{u} d u = \int \frac{4}{x^{2}} d x$

Step 2.2.4

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$\frac{1}{3} (e^{u} + C_{1}) = \int \frac{4}{x^{2}} d x$

Step 2.2.5

Simplify.

$\frac{1}{3} e^{u} + C_{1} = \int \frac{4}{x^{2}} d x$

Step 2.2.6

Replace all occurrences of $u$ with $3 y + 9$ .

$\frac{1}{3} e^{3 y + 9} + C_{1} = \int \frac{4}{x^{2}} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$\frac{1}{3} e^{3 y + 9} + C_{1} = 4 \int \frac{1}{x^{2}} d x$

Step 2.3.2

Apply basic rules of exponents.

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Step 2.3.2.1

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

$\frac{1}{3} e^{3 y + 9} + C_{1} = 4 \int {(x^{2})}^{- 1} d x$

Step 2.3.2.2

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

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Step 2.3.2.2.1

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

$\frac{1}{3} e^{3 y + 9} + C_{1} = 4 \int x^{2 \cdot - 1} d x$

Step 2.3.2.2.2

Multiply $2$ by $- 1$ .

$\frac{1}{3} e^{3 y + 9} + C_{1} = 4 \int x^{- 2} d x$

Step 2.3.3

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

$\frac{1}{3} e^{3 y + 9} + C_{1} = 4 (- x^{- 1} + C_{2})$

Step 2.3.4

Simplify the answer.

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Step 2.3.4.1

Rewrite $4 (- x^{- 1} + C_{2})$ as $4 (- \frac{1}{x}) + C_{2}$ .

$\frac{1}{3} e^{3 y + 9} + C_{1} = 4 (- \frac{1}{x}) + C_{2}$

Step 2.3.4.2

Simplify.

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Step 2.3.4.2.1

Multiply $- 1$ by $4$ .

$\frac{1}{3} e^{3 y + 9} + C_{1} = - 4 \frac{1}{x} + C_{2}$

Step 2.3.4.2.2

Combine $- 4$ and $\frac{1}{x}$ .

$\frac{1}{3} e^{3 y + 9} + C_{1} = \frac{- 4}{x} + C_{2}$

Step 2.3.4.2.3

Move the negative in front of the fraction.

$\frac{1}{3} e^{3 y + 9} + C_{1} = - \frac{4}{x} + C_{2}$

Step 2.4

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

$\frac{1}{3} e^{3 y + 9} = - \frac{4}{x} + K$

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $3$ .

$3 (\frac{1}{3} e^{3 y + 9}) = 3 (- \frac{4}{x} + K)$

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $3 (\frac{1}{3} e^{3 y + 9})$ .

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Step 3.2.1.1.1

Combine $\frac{1}{3}$ and $e^{3 y + 9}$ .

$3 \frac{e^{3 y + 9}}{3} = 3 (- \frac{4}{x} + K)$

Step 3.2.1.1.2

Cancel the common factor of $3$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

$3 \frac{e^{3 y + 9}}{3} = 3 (- \frac{4}{x} + K)$

Step 3.2.1.1.2.2

Rewrite the expression.

$e^{3 y + 9} = 3 (- \frac{4}{x} + K)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $3 (- \frac{4}{x} + K)$ .

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Step 3.2.2.1.1

Apply the distributive property.

$e^{3 y + 9} = 3 (- \frac{4}{x}) + 3 K$

Step 3.2.2.1.2

Multiply $3 (- \frac{4}{x})$ .

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Step 3.2.2.1.2.1

Multiply $- 1$ by $3$ .

$e^{3 y + 9} = - 3 \frac{4}{x} + 3 K$

Step 3.2.2.1.2.2

Combine $- 3$ and $\frac{4}{x}$ .

$e^{3 y + 9} = \frac{- 3 \cdot 4}{x} + 3 K$

Step 3.2.2.1.2.3

Multiply $- 3$ by $4$ .

$e^{3 y + 9} = \frac{- 12}{x} + 3 K$

Step 3.2.2.1.3

Move the negative in front of the fraction.

$e^{3 y + 9} = - \frac{12}{x} + 3 K$

Step 3.3

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

$\ln (e^{3 y + 9}) = \ln (- \frac{12}{x} + 3 K)$

Step 3.4

Expand the left side.

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Step 3.4.1

Expand $\ln (e^{3 y + 9})$ by moving $3 y + 9$ outside the logarithm.

$(3 y + 9) \ln (e) = \ln (- \frac{12}{x} + 3 K)$

Step 3.4.2

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

$(3 y + 9) \cdot 1 = \ln (- \frac{12}{x} + 3 K)$

Step 3.4.3

Multiply $3 y + 9$ by $1$ .

$3 y + 9 = \ln (- \frac{12}{x} + 3 K)$

Step 3.5

Subtract $9$ from both sides of the equation.

$3 y = \ln (- \frac{12}{x} + 3 K) - 9$

Step 3.6

Divide each term in $3 y = \ln (- \frac{12}{x} + 3 K) - 9$ by $3$ and simplify.

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Step 3.6.1

Divide each term in $3 y = \ln (- \frac{12}{x} + 3 K) - 9$ by $3$ .

$\frac{3 y}{3} = \frac{\ln (- \frac{12}{x} + 3 K)}{3} + \frac{- 9}{3}$

Step 3.6.2

Simplify the left side.

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Step 3.6.2.1

Cancel the common factor of $3$ .

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Step 3.6.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{\ln (- \frac{12}{x} + 3 K)}{3} + \frac{- 9}{3}$

Step 3.6.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (- \frac{12}{x} + 3 K)}{3} + \frac{- 9}{3}$

Step 3.6.3

Simplify the right side.

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Step 3.6.3.1

Divide $- 9$ by $3$ .

$y = \frac{\ln (- \frac{12}{x} + 3 K)}{3} - 3$

Step 4

Simplify the constant of integration.

$y = \frac{\ln (- \frac{12}{x} + K)}{3} - 3$