Calculus Examples

$\frac{1}{x^{2} + 5 x + 6}$

Step 1

Rewrite $\frac{1}{x^{2} + 5 x + 6}$ as ${(x^{2} + 5 x + 6)}^{- 1}$ .

$\frac{d}{d x} [{(x^{2} + 5 x + 6)}^{- 1}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2} + 5 x + 6$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $x^{2} + 5 x + 6$ .

$\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{2} + 5 x + 6]$

Step 2.2

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

$- u^{- 2} \frac{d}{d x} [x^{2} + 5 x + 6]$

Step 2.3

Replace all occurrences of $u$ with $x^{2} + 5 x + 6$ .

$- {(x^{2} + 5 x + 6)}^{- 2} \frac{d}{d x} [x^{2} + 5 x + 6]$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $x^{2} + 5 x + 6$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [6]$ .

$- {(x^{2} + 5 x + 6)}^{- 2} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [6])$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$- {(x^{2} + 5 x + 6)}^{- 2} (2 x + \frac{d}{d x} [5 x] + \frac{d}{d x} [6])$

Step 3.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$- {(x^{2} + 5 x + 6)}^{- 2} (2 x + 5 \frac{d}{d x} [x] + \frac{d}{d x} [6])$

Step 3.4

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

$- {(x^{2} + 5 x + 6)}^{- 2} (2 x + 5 \cdot 1 + \frac{d}{d x} [6])$

Step 3.5

Multiply $5$ by $1$ .

$- {(x^{2} + 5 x + 6)}^{- 2} (2 x + 5 + \frac{d}{d x} [6])$

Step 3.6

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

$- {(x^{2} + 5 x + 6)}^{- 2} (2 x + 5 + 0)$

Step 3.7

Add $2 x + 5$ and $0$ .

$- {(x^{2} + 5 x + 6)}^{- 2} (2 x + 5)$

Step 4

Simplify.

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Step 4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{{(x^{2} + 5 x + 6)}^{2}} (2 x + 5)$

Step 4.2

Reorder the factors of $- \frac{1}{{(x^{2} + 5 x + 6)}^{2}} (2 x + 5)$ .

$- (2 x + 5) \frac{1}{{(x^{2} + 5 x + 6)}^{2}}$

Step 4.3

Apply the distributive property.

$(- (2 x) - 1 \cdot 5) \frac{1}{{(x^{2} + 5 x + 6)}^{2}}$

Step 4.4

Multiply $2$ by $- 1$ .

$(- 2 x - 1 \cdot 5) \frac{1}{{(x^{2} + 5 x + 6)}^{2}}$

Step 4.5

Multiply $- 1$ by $5$ .

$(- 2 x - 5) \frac{1}{{(x^{2} + 5 x + 6)}^{2}}$

Step 4.6

Simplify the denominator.

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Step 4.6.1

Factor $x^{2} + 5 x + 6$ using the AC method.

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Step 4.6.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $5$ .

$2, 3$

Step 4.6.1.2

Write the factored form using these integers.

$(- 2 x - 5) \frac{1}{{((x + 2) (x + 3))}^{2}}$

Step 4.6.2

Apply the product rule to $(x + 2) (x + 3)$ .

$(- 2 x - 5) \frac{1}{{(x + 2)}^{2} {(x + 3)}^{2}}$

Step 4.7

Multiply $- 2 x - 5$ by $\frac{1}{{(x + 2)}^{2} {(x + 3)}^{2}}$ .

$\frac{- 2 x - 5}{{(x + 2)}^{2} {(x + 3)}^{2}}$

Step 4.8

Factor $- 1$ out of $- 2 x$ .

$\frac{- (2 x) - 5}{{(x + 2)}^{2} {(x + 3)}^{2}}$

Step 4.9

Rewrite $- 5$ as $- 1 (5)$ .

$\frac{- (2 x) - 1 (5)}{{(x + 2)}^{2} {(x + 3)}^{2}}$

Step 4.10

Factor $- 1$ out of $- (2 x) - 1 (5)$ .

$\frac{- (2 x + 5)}{{(x + 2)}^{2} {(x + 3)}^{2}}$

Step 4.11

Rewrite $- (2 x + 5)$ as $- 1 (2 x + 5)$ .

$\frac{- 1 (2 x + 5)}{{(x + 2)}^{2} {(x + 3)}^{2}}$

Step 4.12

Move the negative in front of the fraction.

$- \frac{2 x + 5}{{(x + 2)}^{2} {(x + 3)}^{2}}$