Calculus Examples

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

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

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

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

$(1 - t^{2}) (0 + \frac{d}{d t} [- \frac{3}{t^{2}}]) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

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

$(1 - t^{2}) (0 - 3 \frac{d}{d t} [\frac{1}{t^{2}}]) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

Step 3.2

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

$(1 - t^{2}) (0 - 3 \frac{d}{d t} [{(t^{2})}^{- 1}]) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

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 $u$ as $t^{2}$ .

$(1 - t^{2}) (0 - 3 (\frac{d}{d u} [u^{- 1}] \frac{d}{d t} [t^{2}])) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

Step 3.3.2

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

$(1 - t^{2}) (0 - 3 (- u^{- 2} \frac{d}{d t} [t^{2}])) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

Step 3.3.3

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

$(1 - t^{2}) (0 - 3 (- {(t^{2})}^{- 2} \frac{d}{d t} [t^{2}])) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

Step 3.4

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

$(1 - t^{2}) (0 - 3 (- {(t^{2})}^{- 2} (2 t))) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.5.2

Multiply $2$ by $- 2$ .

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

Step 3.6

Multiply $2$ by $- 1$ .

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

Subtract $4$ from $1$ .

$(1 - t^{2}) (0 - 3 (- 2 t^{- 3})) + (1 - \frac{3}{t^{2}}) \frac{d}{d t} [1 - t^{2}]$

Step 3.10

Multiply $- 2$ by $- 3$ .

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

Step 4

Simplify.

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Step 4.1

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

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

Step 4.2

Combine terms.

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Step 4.2.1

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

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

Step 4.2.2

Add $0$ and $\frac{6}{t^{3}}$ .

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Multiply $2$ by $- 1$ .

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

Step 7

Subtract $2 t$ from $0$ .

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

Step 8

Simplify.

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Step 8.1

Apply the distributive property.

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

Step 8.2

Apply the distributive property.

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

Step 8.3

Combine terms.

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Step 8.3.1

Multiply $\frac{6}{t^{3}}$ by $1$ .

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

Step 8.3.2

Combine $\frac{6}{t^{3}}$ and $t^{2}$ .

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

Step 8.3.3

Cancel the common factor of $t^{2}$ and $t^{3}$ .

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Step 8.3.3.1

Factor $t^{2}$ out of $6 t^{2}$ .

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

Step 8.3.3.2

Cancel the common factors.

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Step 8.3.3.2.1

Factor $t^{2}$ out of $t^{3}$ .

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

Step 8.3.3.2.2

Cancel the common factor.

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

Step 8.3.3.2.3

Rewrite the expression.

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

Step 8.3.4

Multiply $- 2$ by $1$ .

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

Step 8.3.5

Multiply $- 2$ by $- 1$ .

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

Step 8.3.6

Combine $2$ and $\frac{3}{t^{2}}$ .

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

Step 8.3.7

Multiply $2$ by $3$ .

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

Step 8.3.8

Combine $\frac{6}{t^{2}}$ and $t$ .

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

Step 8.3.9

Cancel the common factor of $t$ and $t^{2}$ .

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Step 8.3.9.1

Factor $t$ out of $6 t$ .

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

Step 8.3.9.2

Cancel the common factors.

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Step 8.3.9.2.1

Factor $t$ out of $t^{2}$ .

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

Step 8.3.9.2.2

Cancel the common factor.

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

Step 8.3.9.2.3

Rewrite the expression.

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

Step 8.3.10

Add $- \frac{6}{t}$ and $\frac{6}{t}$ .

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

Step 8.3.11

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

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

Step 8.4

Reorder terms.

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