Calculus Examples

$s = (\frac{1}{3}) t^{3} - 7 t^{2} + 49 t + 5$

Step 1

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

$s = \frac{1}{3} t^{3} - 7 t^{2} + 49 t + 5$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d t} (s) = \frac{d}{d t} (\frac{1}{3} t^{3} - 7 t^{2} + 49 t + 5)$

Step 3

The derivative of $s$ with respect to $t$ is $s'$ .

$s'$

Step 4

Differentiate the right side of the equation.

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Step 4.1

By the Sum Rule, the derivative of $\frac{1}{3} t^{3} - 7 t^{2} + 49 t + 5$ with respect to $t$ is $\frac{d}{d t} [\frac{1}{3} t^{3}] + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$ .

$\frac{d}{d t} [\frac{1}{3} t^{3}] + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.2

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

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Step 4.2.1

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

$\frac{1}{3} \frac{d}{d t} [t^{3}] + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.2.2

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

$\frac{1}{3} (3 t^{2}) + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.2.3

Combine $3$ and $\frac{1}{3}$ .

$\frac{3}{3} t^{2} + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.2.4

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

$\frac{3 t^{2}}{3} + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.2.5

Cancel the common factor of $3$ .

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Step 4.2.5.1

Cancel the common factor.

$\frac{3 t^{2}}{3} + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.2.5.2

Divide $t^{2}$ by $1$ .

$t^{2} + \frac{d}{d t} [- 7 t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.3

Evaluate $\frac{d}{d t} [- 7 t^{2}]$ .

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Step 4.3.1

Since $- 7$ is constant with respect to $t$ , the derivative of $- 7 t^{2}$ with respect to $t$ is $- 7 \frac{d}{d t} [t^{2}]$ .

$t^{2} - 7 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.3.2

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

$t^{2} - 7 (2 t) + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.3.3

Multiply $2$ by $- 7$ .

$t^{2} - 14 t + \frac{d}{d t} [49 t] + \frac{d}{d t} [5]$

Step 4.4

Evaluate $\frac{d}{d t} [49 t]$ .

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Step 4.4.1

Since $49$ is constant with respect to $t$ , the derivative of $49 t$ with respect to $t$ is $49 \frac{d}{d t} [t]$ .

$t^{2} - 14 t + 49 \frac{d}{d t} [t] + \frac{d}{d t} [5]$

Step 4.4.2

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

$t^{2} - 14 t + 49 \cdot 1 + \frac{d}{d t} [5]$

Step 4.4.3

Multiply $49$ by $1$ .

$t^{2} - 14 t + 49 + \frac{d}{d t} [5]$

Step 4.5

Differentiate using the Constant Rule.

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Step 4.5.1

Since $5$ is constant with respect to $t$ , the derivative of $5$ with respect to $t$ is $0$ .

$t^{2} - 14 t + 49 + 0$

Step 4.5.2

Add $t^{2} - 14 t + 49$ and $0$ .

$t^{2} - 14 t + 49$

Step 5

Reform the equation by setting the left side equal to the right side.

$s' = t^{2} - 14 t + 49$

Step 6

Replace $s'$ with $\frac{d s}{d t}$ .

$\frac{d s}{d t} = t^{2} - 14 t + 49$