Calculus Examples

$\frac{d y}{d x} = \frac{4}{x - 6}$ and $y (7) = - 2$ ?

Step 1

Rewrite the equation.

$d y = \frac{4}{x - 6} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{4}{x - 6} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{4}{x - 6} d x$

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.

$y + C_{1} = 4 \int \frac{1}{x - 6} d x$

Step 2.3.2

Let $u = x - 6$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = x - 6$ . Find $\frac{d u}{d x}$ .

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Step 2.3.2.1.1

Differentiate $x - 6$ .

$\frac{d}{d x} [x - 6]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $x - 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 6]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 6]$

Step 2.3.2.1.3

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

$1 + \frac{d}{d x} [- 6]$

Step 2.3.2.1.4

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.3.2.1.5

Add $1$ and $0$ .

$1$

Step 2.3.2.2

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

$y + C_{1} = 4 \int \frac{1}{u} d u$

Step 2.3.3

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

$y + C_{1} = 4 (\ln (| u |) + C_{2})$

Step 2.3.4

Simplify.

$y + C_{1} = 4 \ln (| u |) + C_{2}$

Step 2.3.5

Replace all occurrences of $u$ with $x - 6$ .

$y + C_{1} = 4 \ln (| x - 6 |) + C_{2}$

Step 2.4

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

$y = 4 \ln (| x - 6 |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $7$ for $x$ and $- 2$ for $y$ in $y = 4 \ln (| x - 6 |) + C$ .

$- 2 = 4 \ln (| 7 - 6 |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $4 \ln (| 7 - 6 |) + C = - 2$ .

$4 \ln (| 7 - 6 |) + C = - 2$

Step 4.2

Simplify $4 \ln (| 7 - 6 |) + C$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Subtract $6$ from $7$ .

$4 \ln (| 1 |) + C = - 2$

Step 4.2.1.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$4 \ln (1) + C = - 2$

Step 4.2.1.3

The natural logarithm of $1$ is $0$ .

$4 \cdot 0 + C = - 2$

Step 4.2.1.4

Multiply $4$ by $0$ .

$0 + C = - 2$

Step 4.2.2

Add $0$ and $C$ .

$C = - 2$

Step 5

Substitute $- 2$ for $C$ in $y = 4 \ln (| x - 6 |) + C$ and simplify.

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Step 5.1

Substitute $- 2$ for $C$ .

$y = 4 \ln (| x - 6 |) - 2$

Step 5.2

Simplify each term.

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Step 5.2.1

Simplify $4 \ln (| x - 6 |)$ by moving $4$ inside the logarithm.

$y = \ln ({| x - 6 |}^{4}) - 2$

Step 5.2.2

Remove the absolute value in ${| x - 6 |}^{4}$ because exponentiations with even powers are always positive.

$y = \ln ({(x - 6)}^{4}) - 2$