Calculus Examples

$P (t) = t^{4} + 5 t^{3} - 2 t^{2} + \frac{6}{7 t + 4}$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $t^{4} + 5 t^{3} - 2 t^{2} + \frac{6}{7 t + 4}$ with respect to $t$ is $\frac{d}{d t} [t^{4}] + \frac{d}{d t} [5 t^{3}] + \frac{d}{d t} [- 2 t^{2}] + \frac{d}{d t} [\frac{6}{7 t + 4}]$ .

$\frac{d}{d t} [t^{4}] + \frac{d}{d t} [5 t^{3}] + \frac{d}{d t} [- 2 t^{2}] + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 4$ .

$4 t^{3} + \frac{d}{d t} [5 t^{3}] + \frac{d}{d t} [- 2 t^{2}] + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 2

Evaluate $\frac{d}{d t} [5 t^{3}]$ .

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Step 2.1

Since $5$ is constant with respect to $t$ , the derivative of $5 t^{3}$ with respect to $t$ is $5 \frac{d}{d t} [t^{3}]$ .

$4 t^{3} + 5 \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 2 t^{2}] + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 2.2

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

$4 t^{3} + 5 (3 t^{2}) + \frac{d}{d t} [- 2 t^{2}] + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 2.3

Multiply $3$ by $5$ .

$4 t^{3} + 15 t^{2} + \frac{d}{d t} [- 2 t^{2}] + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 3

Evaluate $\frac{d}{d t} [- 2 t^{2}]$ .

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Step 3.1

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

$4 t^{3} + 15 t^{2} - 2 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 3.2

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

$4 t^{3} + 15 t^{2} - 2 (2 t) + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 3.3

Multiply $2$ by $- 2$ .

$4 t^{3} + 15 t^{2} - 4 t + \frac{d}{d t} [\frac{6}{7 t + 4}]$

Step 4

Evaluate $\frac{d}{d t} [\frac{6}{7 t + 4}]$ .

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Step 4.1

Since $6$ is constant with respect to $t$ , the derivative of $\frac{6}{7 t + 4}$ with respect to $t$ is $6 \frac{d}{d t} [\frac{1}{7 t + 4}]$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 \frac{d}{d t} [\frac{1}{7 t + 4}]$

Step 4.2

Rewrite $\frac{1}{7 t + 4}$ as ${(7 t + 4)}^{- 1}$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 \frac{d}{d t} [{(7 t + 4)}^{- 1}]$

Step 4.3

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 $g (t) = 7 t + 4$ .

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Step 4.3.1

To apply the Chain Rule, set $u$ as $7 t + 4$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (\frac{d}{d u} [u^{- 1}] \frac{d}{d t} [7 t + 4])$

Step 4.3.2

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

$4 t^{3} + 15 t^{2} - 4 t + 6 (- u^{- 2} \frac{d}{d t} [7 t + 4])$

Step 4.3.3

Replace all occurrences of $u$ with $7 t + 4$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (- {(7 t + 4)}^{- 2} \frac{d}{d t} [7 t + 4])$

Step 4.4

By the Sum Rule, the derivative of $7 t + 4$ with respect to $t$ is $\frac{d}{d t} [7 t] + \frac{d}{d t} [4]$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (- {(7 t + 4)}^{- 2} (\frac{d}{d t} [7 t] + \frac{d}{d t} [4]))$

Step 4.5

Since $7$ is constant with respect to $t$ , the derivative of $7 t$ with respect to $t$ is $7 \frac{d}{d t} [t]$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (- {(7 t + 4)}^{- 2} (7 \frac{d}{d t} [t] + \frac{d}{d t} [4]))$

Step 4.6

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

$4 t^{3} + 15 t^{2} - 4 t + 6 (- {(7 t + 4)}^{- 2} (7 \cdot 1 + \frac{d}{d t} [4]))$

Step 4.7

Since $4$ is constant with respect to $t$ , the derivative of $4$ with respect to $t$ is $0$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (- {(7 t + 4)}^{- 2} (7 \cdot 1 + 0))$

Step 4.8

Multiply $7$ by $1$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (- {(7 t + 4)}^{- 2} (7 + 0))$

Step 4.9

Add $7$ and $0$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (- {(7 t + 4)}^{- 2} \cdot 7)$

Step 4.10

Multiply $7$ by $- 1$ .

$4 t^{3} + 15 t^{2} - 4 t + 6 (- 7 {(7 t + 4)}^{- 2})$

Step 4.11

Multiply $- 7$ by $6$ .

$4 t^{3} + 15 t^{2} - 4 t - 42 {(7 t + 4)}^{- 2}$

Step 5

Simplify.

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Step 5.1

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

$4 t^{3} + 15 t^{2} - 4 t - 42 \frac{1}{{(7 t + 4)}^{2}}$

Step 5.2

Combine terms.

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Step 5.2.1

Combine $- 42$ and $\frac{1}{{(7 t + 4)}^{2}}$ .

$4 t^{3} + 15 t^{2} - 4 t + \frac{- 42}{{(7 t + 4)}^{2}}$

Step 5.2.2

Move the negative in front of the fraction.

$4 t^{3} + 15 t^{2} - 4 t - \frac{42}{{(7 t + 4)}^{2}}$