Calculus Examples

$\frac{6}{t^{2} + 5 t - 9}$

Step 1

Differentiate using the Constant Multiple Rule.

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Step 1.1

Since $6$ is constant with respect to $t$ , the derivative of $\frac{6}{t^{2} + 5 t - 9}$ with respect to $t$ is $6 \frac{d}{d t} [\frac{1}{t^{2} + 5 t - 9}]$ .

$6 \frac{d}{d t} [\frac{1}{t^{2} + 5 t - 9}]$

Step 1.2

Rewrite $\frac{1}{t^{2} + 5 t - 9}$ as ${(t^{2} + 5 t - 9)}^{- 1}$ .

$6 \frac{d}{d t} [{(t^{2} + 5 t - 9)}^{- 1}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = t^{2} + 5 t - 9$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $t^{2} + 5 t - 9$ .

$6 (\frac{d}{d u} [u^{- 1}] \frac{d}{d t} [t^{2} + 5 t - 9])$

Step 2.2

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

$6 (- u^{- 2} \frac{d}{d t} [t^{2} + 5 t - 9])$

Step 2.3

Replace all occurrences of $u$ with $t^{2} + 5 t - 9$ .

$6 (- {(t^{2} + 5 t - 9)}^{- 2} \frac{d}{d t} [t^{2} + 5 t - 9])$

Step 3

Differentiate.

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Step 3.1

Multiply $- 1$ by $6$ .

$- 6 ({(t^{2} + 5 t - 9)}^{- 2} \frac{d}{d t} [t^{2} + 5 t - 9])$

Step 3.2

By the Sum Rule, the derivative of $t^{2} + 5 t - 9$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [5 t] + \frac{d}{d t} [- 9]$ .

$- 6 {(t^{2} + 5 t - 9)}^{- 2} (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [5 t] + \frac{d}{d t} [- 9])$

Step 3.3

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

$- 6 {(t^{2} + 5 t - 9)}^{- 2} (2 t + \frac{d}{d t} [5 t] + \frac{d}{d t} [- 9])$

Step 3.4

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

$- 6 {(t^{2} + 5 t - 9)}^{- 2} (2 t + 5 \frac{d}{d t} [t] + \frac{d}{d t} [- 9])$

Step 3.5

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

$- 6 {(t^{2} + 5 t - 9)}^{- 2} (2 t + 5 \cdot 1 + \frac{d}{d t} [- 9])$

Step 3.6

Multiply $5$ by $1$ .

$- 6 {(t^{2} + 5 t - 9)}^{- 2} (2 t + 5 + \frac{d}{d t} [- 9])$

Step 3.7

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

$- 6 {(t^{2} + 5 t - 9)}^{- 2} (2 t + 5 + 0)$

Step 3.8

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

$- 6 {(t^{2} + 5 t - 9)}^{- 2} (2 t + 5)$

Step 4

Simplify.

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Step 4.1

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

$- 6 \frac{1}{{(t^{2} + 5 t - 9)}^{2}} (2 t + 5)$

Step 4.2

Combine terms.

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Step 4.2.1

Combine $- 6$ and $\frac{1}{{(t^{2} + 5 t - 9)}^{2}}$ .

$\frac{- 6}{{(t^{2} + 5 t - 9)}^{2}} (2 t + 5)$

Step 4.2.2

Move the negative in front of the fraction.

$- \frac{6}{{(t^{2} + 5 t - 9)}^{2}} (2 t + 5)$

Step 4.3

Reorder the factors of $- \frac{6}{{(t^{2} + 5 t - 9)}^{2}} (2 t + 5)$ .

$- (2 t + 5) \frac{6}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.4

Apply the distributive property.

$(- (2 t) - 1 \cdot 5) \frac{6}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.5

Multiply $2$ by $- 1$ .

$(- 2 t - 1 \cdot 5) \frac{6}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.6

Multiply $- 1$ by $5$ .

$(- 2 t - 5) \frac{6}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.7

Multiply $- 2 t - 5$ by $\frac{6}{{(t^{2} + 5 t - 9)}^{2}}$ .

$\frac{(- 2 t - 5) \cdot 6}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.8

Move $6$ to the left of $- 2 t - 5$ .

$\frac{6 \cdot (- 2 t - 5)}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.9

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

$\frac{6 (- (2 t) - 5)}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.10

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

$\frac{6 (- (2 t) - 1 (5))}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.11

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

$\frac{6 (- (2 t + 5))}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.12

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

$\frac{6 (- 1 (2 t + 5))}{{(t^{2} + 5 t - 9)}^{2}}$

Step 4.13

Move the negative in front of the fraction.

$- \frac{6 (2 t + 5)}{{(t^{2} + 5 t - 9)}^{2}}$