Calculus Examples

${(1 - \frac{t}{40})}^{2}$

Step 1

Rewrite ${(1 - \frac{t}{40})}^{2}$ as $(1 - \frac{t}{40}) (1 - \frac{t}{40})$ .

$\frac{d}{d t} [(1 - \frac{t}{40}) (1 - \frac{t}{40})]$

Step 2

Expand $(1 - \frac{t}{40}) (1 - \frac{t}{40})$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

$\frac{d}{d t} [1 (1 - \frac{t}{40}) - \frac{t}{40} (1 - \frac{t}{40})]$

Step 2.2

Apply the distributive property.

$\frac{d}{d t} [1 \cdot 1 + 1 (- \frac{t}{40}) - \frac{t}{40} (1 - \frac{t}{40})]$

Step 2.3

Apply the distributive property.

$\frac{d}{d t} [1 \cdot 1 + 1 (- \frac{t}{40}) - \frac{t}{40} \cdot 1 - \frac{t}{40} (- \frac{t}{40})]$

Step 3

Simplify and combine like terms.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply $1$ by $1$ .

$\frac{d}{d t} [1 + 1 (- \frac{t}{40}) - \frac{t}{40} \cdot 1 - \frac{t}{40} (- \frac{t}{40})]$

Step 3.1.2

Multiply $- \frac{t}{40}$ by $1$ .

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} \cdot 1 - \frac{t}{40} (- \frac{t}{40})]$

Step 3.1.3

Multiply $- 1$ by $1$ .

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} - \frac{t}{40} (- \frac{t}{40})]$

Step 3.1.4

Multiply $- \frac{t}{40} (- \frac{t}{40})$ .

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Step 3.1.4.1

Multiply $- 1$ by $- 1$ .

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + 1 \frac{t}{40} \frac{t}{40}]$

Step 3.1.4.2

Multiply $\frac{t}{40}$ by $1$ .

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + \frac{t}{40} \cdot \frac{t}{40}]$

Step 3.1.4.3

Multiply $\frac{t}{40}$ by $\frac{t}{40}$ .

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + \frac{t \cdot t}{40 \cdot 40}]$

Step 3.1.4.4

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

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + \frac{t^{1} t}{40 \cdot 40}]$

Step 3.1.4.5

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

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + \frac{t^{1} t^{1}}{40 \cdot 40}]$

Step 3.1.4.6

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

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + \frac{t^{1 + 1}}{40 \cdot 40}]$

Step 3.1.4.7

Add $1$ and $1$ .

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + \frac{t^{2}}{40 \cdot 40}]$

Step 3.1.4.8

Multiply $40$ by $40$ .

$\frac{d}{d t} [1 - \frac{t}{40} - \frac{t}{40} + \frac{t^{2}}{1600}]$

Step 3.2

Subtract $\frac{t}{40}$ from $- \frac{t}{40}$ .

$\frac{d}{d t} [1 - 2 \frac{t}{40} + \frac{t^{2}}{1600}]$

Step 4

Simplify each term.

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Step 4.1

Cancel the common factor of $2$ .

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Step 4.1.1

Factor $2$ out of $- 2$ .

$\frac{d}{d t} [1 + 2 (- 1) \frac{t}{40} + \frac{t^{2}}{1600}]$

Step 4.1.2

Factor $2$ out of $40$ .

$\frac{d}{d t} [1 + 2 \cdot - 1 \frac{t}{2 \cdot 20} + \frac{t^{2}}{1600}]$

Step 4.1.3

Cancel the common factor.

$\frac{d}{d t} [1 + 2 \cdot - 1 \frac{t}{2 \cdot 20} + \frac{t^{2}}{1600}]$

Step 4.1.4

Rewrite the expression.

$\frac{d}{d t} [1 - 1 \frac{t}{20} + \frac{t^{2}}{1600}]$

Step 4.2

Rewrite $- 1 \frac{t}{20}$ as $- \frac{t}{20}$ .

$\frac{d}{d t} [1 - \frac{t}{20} + \frac{t^{2}}{1600}]$

Step 5

By the Sum Rule, the derivative of $1 - \frac{t}{20} + \frac{t^{2}}{1600}$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [- \frac{t}{20}] + \frac{d}{d t} [\frac{t^{2}}{1600}]$ .

$\frac{d}{d t} [1] + \frac{d}{d t} [- \frac{t}{20}] + \frac{d}{d t} [\frac{t^{2}}{1600}]$

Step 6

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

$0 + \frac{d}{d t} [- \frac{t}{20}] + \frac{d}{d t} [\frac{t^{2}}{1600}]$

Step 7

Add $0$ and $\frac{d}{d t} [- \frac{t}{20}]$ .

$\frac{d}{d t} [- \frac{t}{20}] + \frac{d}{d t} [\frac{t^{2}}{1600}]$

Step 8

Since $- \frac{1}{20}$ is constant with respect to $t$ , the derivative of $- \frac{t}{20}$ with respect to $t$ is $- \frac{1}{20} \frac{d}{d t} [t]$ .

$- \frac{1}{20} \frac{d}{d t} [t] + \frac{d}{d t} [\frac{t^{2}}{1600}]$

Step 9

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

$- \frac{1}{20} \cdot 1 + \frac{d}{d t} [\frac{t^{2}}{1600}]$

Step 10

Multiply $- 1$ by $1$ .

$- \frac{1}{20} + \frac{d}{d t} [\frac{t^{2}}{1600}]$

Step 11

Since $\frac{1}{1600}$ is constant with respect to $t$ , the derivative of $\frac{t^{2}}{1600}$ with respect to $t$ is $\frac{1}{1600} \frac{d}{d t} [t^{2}]$ .

$- \frac{1}{20} + \frac{1}{1600} \frac{d}{d t} [t^{2}]$

Step 12

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

$- \frac{1}{20} + \frac{1}{1600} (2 t)$

Step 13

Simplify terms.

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Step 13.1

Combine $2$ and $\frac{1}{1600}$ .

$- \frac{1}{20} + \frac{2}{1600} t$

Step 13.2

Combine $\frac{2}{1600}$ and $t$ .

$- \frac{1}{20} + \frac{2 t}{1600}$

Step 13.3

Cancel the common factor of $2$ and $1600$ .

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Step 13.3.1

Factor $2$ out of $2 t$ .

$- \frac{1}{20} + \frac{2 (t)}{1600}$

Step 13.3.2

Cancel the common factors.

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Step 13.3.2.1

Factor $2$ out of $1600$ .

$- \frac{1}{20} + \frac{2 t}{2 \cdot 800}$

Step 13.3.2.2

Cancel the common factor.

$- \frac{1}{20} + \frac{2 t}{2 \cdot 800}$

Step 13.3.2.3

Rewrite the expression.

$- \frac{1}{20} + \frac{t}{800}$

Step 13.4

Reorder terms.

$\frac{t}{800} - \frac{1}{20}$