Precalculus Examples

$n (t) = \frac{1}{t} + \frac{3}{t^{2}} + \frac{2}{t^{3}} + 4$

Step 1

By the Sum Rule, the derivative of $\frac{1}{t} + \frac{3}{t^{2}} + \frac{2}{t^{3}} + 4$ with respect to $t$ is $\frac{d}{d t} [\frac{1}{t}] + \frac{d}{d t} [\frac{3}{t^{2}}] + \frac{d}{d t} [\frac{2}{t^{3}}] + \frac{d}{d t} [4]$ .

$\frac{d}{d t} [\frac{1}{t}] + \frac{d}{d t} [\frac{3}{t^{2}}] + \frac{d}{d t} [\frac{2}{t^{3}}] + \frac{d}{d t} [4]$

Step 2

Evaluate $\frac{d}{d t} [\frac{1}{t}]$ .

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Step 2.1

Rewrite $\frac{1}{t}$ as $t^{- 1}$ .

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

Step 2.2

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

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

Step 3

Evaluate $\frac{d}{d t} [\frac{3}{t^{2}}]$ .

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Step 3.1

Since $3$ is constant with respect to $t$ , the derivative of $\frac{3}{t^{2}}$ with respect to $t$ is $3 \frac{d}{d t} [\frac{1}{t^{2}}]$ .

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

Step 3.2

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

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

Step 3.3

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 $g (t) = t^{2}$ .

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Step 3.3.1

To apply the Chain Rule, set $u_{1}$ as $t^{2}$ .

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

Step 3.3.2

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

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

Step 3.3.3

Replace all occurrences of $u_{1}$ with $t^{2}$ .

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

Step 3.4

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

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

Step 3.5

Multiply the exponents in ${(t^{2})}^{- 2}$ .

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Step 3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.5.2

Multiply $2$ by $- 2$ .

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

Step 3.6

Multiply $2$ by $- 1$ .

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

Step 3.7

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

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

Step 3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.9

Subtract $4$ from $1$ .

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

Step 3.10

Multiply $- 2$ by $3$ .

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

Step 4

Evaluate $\frac{d}{d t} [\frac{2}{t^{3}}]$ .

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Step 4.1

Since $2$ is constant with respect to $t$ , the derivative of $\frac{2}{t^{3}}$ with respect to $t$ is $2 \frac{d}{d t} [\frac{1}{t^{3}}]$ .

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

Step 4.2

Rewrite $\frac{1}{t^{3}}$ as ${(t^{3})}^{- 1}$ .

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

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 $g (t) = t^{3}$ .

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Step 4.3.1

To apply the Chain Rule, set $u_{2}$ as $t^{3}$ .

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

Step 4.3.2

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

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

Step 4.3.3

Replace all occurrences of $u_{2}$ with $t^{3}$ .

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

Step 4.4

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

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

Step 4.5

Multiply the exponents in ${(t^{3})}^{- 2}$ .

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Step 4.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 4.5.2

Multiply $3$ by $- 2$ .

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

Step 4.6

Multiply $3$ by $- 1$ .

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

Step 4.7

Multiply $t^{- 6}$ by $t^{2}$ by adding the exponents.

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Step 4.7.1

Move $t^{2}$ .

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

Step 4.7.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 4.7.3

Subtract $6$ from $2$ .

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

Step 4.8

Multiply $- 3$ by $2$ .

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

Step 5

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

$- t^{- 2} - 6 t^{- 3} - 6 t^{- 4} + 0$

Step 6

Simplify.

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Step 6.1

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

$- \frac{1}{t^{2}} - 6 t^{- 3} - 6 t^{- 4} + 0$

Step 6.2

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

$- \frac{1}{t^{2}} - 6 \frac{1}{t^{3}} - 6 t^{- 4} + 0$

Step 6.3

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

$- \frac{1}{t^{2}} - 6 \frac{1}{t^{3}} - 6 \frac{1}{t^{4}} + 0$

Step 6.4

Combine terms.

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Step 6.4.1

Combine $- 6$ and $\frac{1}{t^{3}}$ .

$- \frac{1}{t^{2}} + \frac{- 6}{t^{3}} - 6 \frac{1}{t^{4}} + 0$

Step 6.4.2

Move the negative in front of the fraction.

$- \frac{1}{t^{2}} - \frac{6}{t^{3}} - 6 \frac{1}{t^{4}} + 0$

Step 6.4.3

Combine $- 6$ and $\frac{1}{t^{4}}$ .

$- \frac{1}{t^{2}} - \frac{6}{t^{3}} + \frac{- 6}{t^{4}} + 0$

Step 6.4.4

Move the negative in front of the fraction.

$- \frac{1}{t^{2}} - \frac{6}{t^{3}} - \frac{6}{t^{4}} + 0$

Step 6.4.5

Add $- \frac{1}{t^{2}} - \frac{6}{t^{3}} - \frac{6}{t^{4}}$ and $0$ .

$- \frac{1}{t^{2}} - \frac{6}{t^{3}} - \frac{6}{t^{4}}$