Calculus Examples

$S (t) = 2 t^{\frac{3}{2}} - 9 t + 6$

Step 1

Find the first derivative.

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Step 1.1

By the Sum Rule, the derivative of $2 t^{\frac{3}{2}} - 9 t + 6$ with respect to $t$ is $\frac{d}{d t} [2 t^{\frac{3}{2}}] + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$ .

$\frac{d}{d t} [2 t^{\frac{3}{2}}] + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2

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

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Step 1.2.1

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

$2 \frac{d}{d t} [t^{\frac{3}{2}}] + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = \frac{3}{2}$ .

$2 (\frac{3}{2} t^{\frac{3}{2} - 1}) + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 (\frac{3}{2} t^{\frac{3}{2} - 1 \cdot \frac{2}{2}}) + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.4

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

$2 (\frac{3}{2} t^{\frac{3}{2} + \frac{- 1 \cdot 2}{2}}) + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.5

Combine the numerators over the common denominator.

$2 (\frac{3}{2} t^{\frac{3 - 1 \cdot 2}{2}}) + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.6

Simplify the numerator.

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Step 1.2.6.1

Multiply $- 1$ by $2$ .

$2 (\frac{3}{2} t^{\frac{3 - 2}{2}}) + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.6.2

Subtract $2$ from $3$ .

$2 (\frac{3}{2} t^{\frac{1}{2}}) + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.7

Combine $\frac{3}{2}$ and $t^{\frac{1}{2}}$ .

$2 \frac{3 t^{\frac{1}{2}}}{2} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.8

Combine $2$ and $\frac{3 t^{\frac{1}{2}}}{2}$ .

$\frac{2 (3 t^{\frac{1}{2}})}{2} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.9

Multiply $3$ by $2$ .

$\frac{6 t^{\frac{1}{2}}}{2} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.10

Factor $2$ out of $6 t^{\frac{1}{2}}$ .

$\frac{2 (3 t^{\frac{1}{2}})}{2} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.11

Cancel the common factors.

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Step 1.2.11.1

Factor $2$ out of $2$ .

$\frac{2 (3 t^{\frac{1}{2}})}{2 (1)} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.11.2

Cancel the common factor.

$\frac{2 (3 t^{\frac{1}{2}})}{2 \cdot 1} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.11.3

Rewrite the expression.

$\frac{3 t^{\frac{1}{2}}}{1} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.2.11.4

Divide $3 t^{\frac{1}{2}}$ by $1$ .

$3 t^{\frac{1}{2}} + \frac{d}{d t} [- 9 t] + \frac{d}{d t} [6]$

Step 1.3

Evaluate $\frac{d}{d t} [- 9 t]$ .

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Step 1.3.1

Since $- 9$ is constant with respect to $t$ , the derivative of $- 9 t$ with respect to $t$ is $- 9 \frac{d}{d t} [t]$ .

$3 t^{\frac{1}{2}} - 9 \frac{d}{d t} [t] + \frac{d}{d t} [6]$

Step 1.3.2

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

$3 t^{\frac{1}{2}} - 9 \cdot 1 + \frac{d}{d t} [6]$

Step 1.3.3

Multiply $- 9$ by $1$ .

$3 t^{\frac{1}{2}} - 9 + \frac{d}{d t} [6]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $6$ is constant with respect to $t$ , the derivative of $6$ with respect to $t$ is $0$ .

$3 t^{\frac{1}{2}} - 9 + 0$

Step 1.4.2

Add $3 t^{\frac{1}{2}} - 9$ and $0$ .

$f' (t) = 3 t^{\frac{1}{2}} - 9$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $3 t^{\frac{1}{2}} - 9$ with respect to $t$ is $\frac{d}{d t} [3 t^{\frac{1}{2}}] + \frac{d}{d t} [- 9]$ .

$\frac{d}{d t} [3 t^{\frac{1}{2}}] + \frac{d}{d t} [- 9]$

Step 2.2

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

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Step 2.2.1

Since $3$ is constant with respect to $t$ , the derivative of $3 t^{\frac{1}{2}}$ with respect to $t$ is $3 \frac{d}{d t} [t^{\frac{1}{2}}]$ .

$3 \frac{d}{d t} [t^{\frac{1}{2}}] + \frac{d}{d t} [- 9]$

Step 2.2.2

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

$3 (\frac{1}{2} t^{\frac{1}{2} - 1}) + \frac{d}{d t} [- 9]$

Step 2.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$3 (\frac{1}{2} t^{\frac{1}{2} - 1 \cdot \frac{2}{2}}) + \frac{d}{d t} [- 9]$

Step 2.2.4

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

$3 (\frac{1}{2} t^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}}) + \frac{d}{d t} [- 9]$

Step 2.2.5

Combine the numerators over the common denominator.

$3 (\frac{1}{2} t^{\frac{1 - 1 \cdot 2}{2}}) + \frac{d}{d t} [- 9]$

Step 2.2.6

Simplify the numerator.

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Step 2.2.6.1

Multiply $- 1$ by $2$ .

$3 (\frac{1}{2} t^{\frac{1 - 2}{2}}) + \frac{d}{d t} [- 9]$

Step 2.2.6.2

Subtract $2$ from $1$ .

$3 (\frac{1}{2} t^{\frac{- 1}{2}}) + \frac{d}{d t} [- 9]$

Step 2.2.7

Move the negative in front of the fraction.

$3 (\frac{1}{2} t^{- \frac{1}{2}}) + \frac{d}{d t} [- 9]$

Step 2.2.8

Combine $\frac{1}{2}$ and $t^{- \frac{1}{2}}$ .

$3 \frac{t^{- \frac{1}{2}}}{2} + \frac{d}{d t} [- 9]$

Step 2.2.9

Combine $3$ and $\frac{t^{- \frac{1}{2}}}{2}$ .

$\frac{3 t^{- \frac{1}{2}}}{2} + \frac{d}{d t} [- 9]$

Step 2.2.10

Move $t^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{3}{2 t^{\frac{1}{2}}} + \frac{d}{d t} [- 9]$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $- 9$ is constant with respect to $t$ , the derivative of $- 9$ with respect to $t$ is $0$ .

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

Step 2.3.2

Add $\frac{3}{2 t^{\frac{1}{2}}}$ and $0$ .

$f'' (t) = \frac{3}{2 t^{\frac{1}{2}}}$

Step 3

The second derivative of $S (t)$ with respect to $t$ is $\frac{3}{2 t^{\frac{1}{2}}}$ .

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