Calculus Examples

$t^{- 3} (2 t - 2)$

Step 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 $f (t) = t^{- 3}$ and $g (t) = 2 t - 2$ .

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Multiply $2$ by $1$ .

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

Step 2.5

Since $- 2$ is constant with respect to $t$ , the derivative of $- 2$ with respect to $t$ is $0$ .

$t^{- 3} (2 + 0) + (2 t - 2) \frac{d}{d t} [t^{- 3}]$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $2$ and $0$ .

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

Step 2.6.2

Move $2$ to the left of $t^{- 3}$ .

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

Step 2.7

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

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

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

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

Step 3.2

Combine terms.

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Step 3.2.1

Multiply $- 3$ by $2$ .

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

Step 3.2.2

Multiply $t$ by $t^{- 4}$ by adding the exponents.

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Step 3.2.2.1

Move $t^{- 4}$ .

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

Step 3.2.2.2

Multiply $t^{- 4}$ by $t$ .

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Step 3.2.2.2.1

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

$2 t^{- 3} - 6 (t^{- 4} t^{1}) - 2 (- 3 t^{- 4})$

Step 3.2.2.2.2

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

$2 t^{- 3} - 6 t^{- 4 + 1} - 2 (- 3 t^{- 4})$

Step 3.2.2.3

Add $- 4$ and $1$ .

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

Step 3.2.3

Multiply $- 3$ by $- 2$ .

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

Step 3.2.4

Subtract $6 t^{- 3}$ from $2 t^{- 3}$ .

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

Step 3.3

Simplify each term.

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Move the negative in front of the fraction.

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

Step 3.3.4

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

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

Step 3.3.5

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

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