Find the Derivative minus d First, divide dx y equals(x to the power of three plus eight) divide(x plus two)

$y = \frac{x^{3} + 8}{x + 2}$

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{3} + 8$ and $g (x) = x + 2$ .

$\frac{(x + 2) \frac{d}{d x} [x^{3} + 8] - (x^{3} + 8) \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Differentiate.

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By the Sum Rule, the derivative of $x^{3} + 8$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [8]$ .

$\frac{(x + 2) (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [8]) - (x^{3} + 8) \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$\frac{(x + 2) (3 x^{2} + \frac{d}{d x} [8]) - (x^{3} + 8) \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$\frac{(x + 2) (3 x^{2} + 0) - (x^{3} + 8) \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Clarify the expression.

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Add $3 x^{2}$ and $0$ .

$\frac{(x + 2) (3 x^{2}) - (x^{3} + 8) \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Put $3$ to the left of $x + 2$ .

$\frac{3 \cdot (x + 2) x^{2} - (x^{3} + 8) \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

By the Sum Rule, the derivative of $x + 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [2]$ .

$\frac{3 (x + 2) x^{2} - (x^{3} + 8) (\frac{d}{d x} [x] + \frac{d}{d x} [2])}{{(x + 2)}^{2}}$

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

$\frac{3 (x + 2) x^{2} - (x^{3} + 8) (1 + \frac{d}{d x} [2])}{{(x + 2)}^{2}}$

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{3 (x + 2) x^{2} - (x^{3} + 8) (1 + 0)}{{(x + 2)}^{2}}$

Clarify the expression.

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Add $1$ and $0$ .

$\frac{3 (x + 2) x^{2} - (x^{3} + 8) \cdot 1}{{(x + 2)}^{2}}$

First, multiply $- 1$ by $1$ .

$\frac{3 (x + 2) x^{2} - (x^{3} + 8)}{{(x + 2)}^{2}}$

Simplify.

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Use the distributive law.

$\frac{(3 x + 3 \cdot 2) x^{2} - (x^{3} + 8)}{{(x + 2)}^{2}}$

Apply the distributive property.

$\frac{3 x \cdot x^{2} + 3 \cdot 2 x^{2} - (x^{3} + 8)}{{(x + 2)}^{2}}$

Apply the distributive property.

$\frac{3 x \cdot x^{2} + 3 \cdot 2 x^{2} - x^{3} - 1 \cdot 8}{{(x + 2)}^{2}}$

Clarify the numerator.

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Clarify each term.

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Multiply $x$ by $x^{2}$ by adding the exponents.

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Put $x^{2}$ .

$\frac{3 (x^{2} x) + 3 \cdot 2 x^{2} - x^{3} - 1 \cdot 8}{{(x + 2)}^{2}}$

Multiply $x^{2}$ by $x$ .

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Raise $x$ to the power of $1$ .

$\frac{3 (x^{2} x^{1}) + 3 \cdot 2 x^{2} - x^{3} - 1 \cdot 8}{{(x + 2)}^{2}}$

Apply the power rule $a^{m} a^{n} = a^{m + n}$ to Add exponents.

$\frac{3 x^{2 + 1} + 3 \cdot 2 x^{2} - x^{3} - 1 \cdot 8}{{(x + 2)}^{2}}$

Add $2$ and $1$ .

$\frac{3 x^{3} + 3 \cdot 2 x^{2} - x^{3} - 1 \cdot 8}{{(x + 2)}^{2}}$

Multiply $3$ by $2$ .

$\frac{3 x^{3} + 6 x^{2} - x^{3} - 1 \cdot 8}{{(x + 2)}^{2}}$

Multiply $- 1$ by $8$ .

$\frac{3 x^{3} + 6 x^{2} - x^{3} - 8}{{(x + 2)}^{2}}$

Subtract $x^{3}$ from $3 x^{3}$ .

$\frac{2 x^{3} + 6 x^{2} - 8}{{(x + 2)}^{2}}$

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