Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Multiply $3$ by $2$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Multiply $6$ by $1$ .

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

Step 4

Evaluate $\frac{d}{d t} [- \frac{4}{t^{2}}]$ .

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Step 4.1

Since $- 4$ is constant with respect to $t$ , the derivative of $- \frac{4}{t^{2}}$ with respect to $t$ is $- 4 \frac{d}{d t} [\frac{1}{t^{2}}]$ .

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

Step 4.2

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

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

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

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Step 4.3.1

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

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

Step 4.3.2

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

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

Step 4.3.3

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

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

Step 4.4

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

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

Step 4.5

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

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Step 4.5.1

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

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

Step 4.5.2

Multiply $2$ by $- 2$ .

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

Step 4.6

Multiply $2$ by $- 1$ .

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

Step 4.7

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

$6 t^{2} + 6 - 4 (- 2 (t^{1} t^{- 4}))$

Step 4.8

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

$6 t^{2} + 6 - 4 (- 2 t^{1 - 4})$

Step 4.9

Subtract $4$ from $1$ .

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

Step 4.10

Multiply $- 2$ by $- 4$ .

$6 t^{2} + 6 + 8 t^{- 3}$

Step 5

Simplify.

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Step 5.1

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

$6 t^{2} + 6 + 8 \frac{1}{t^{3}}$

Step 5.2

Combine $8$ and $\frac{1}{t^{3}}$ .

$6 t^{2} + 6 + \frac{8}{t^{3}}$

Step 5.3

Reorder terms.

$6 t^{2} + \frac{8}{t^{3}} + 6$