Calculus Examples

$g (t) = t^{2} - \frac{1}{t^{3}}$

Step 1

Differentiate.

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Step 1.1

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

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

Step 1.2

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

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

Step 2

Evaluate $\frac{d}{d t} [- \frac{1}{t^{3}}]$ .

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Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = - 1$ and $g (t) = \frac{1}{t^{3}}$ .

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

Step 2.2

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

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

Step 2.3

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 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

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

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

Step 2.5

Since $- 1$ is constant with respect to $t$ , the derivative of $- 1$ with respect to $t$ is $0$ .

$2 t - (- {(t^{3})}^{- 2} (3 t^{2})) + \frac{1}{t^{3}} \cdot 0$

Step 2.6

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

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Step 2.6.1

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

$2 t - (- t^{3 \cdot - 2} (3 t^{2})) + \frac{1}{t^{3}} \cdot 0$

Step 2.6.2

Multiply $3$ by $- 2$ .

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

Step 2.7

Multiply $3$ by $- 1$ .

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

Step 2.8

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

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Step 2.8.1

Move $t^{2}$ .

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

Step 2.8.2

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

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

Step 2.8.3

Subtract $6$ from $2$ .

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

Step 2.9

Multiply $- 3$ by $- 1$ .

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

Step 2.10

Multiply $\frac{1}{t^{3}}$ by $0$ .

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

Step 2.11

Add $3 t^{- 4}$ and $0$ .

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

Step 3

Simplify.

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Step 3.1

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

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

Step 3.2

Combine $3$ and $\frac{1}{t^{4}}$ .

$2 t + \frac{3}{t^{4}}$