Precalculus Examples

$s (t) = - 1 - 13 t$ , $t = 8$

Step 1

Find the derivative.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $- 1 - 13 t$ with respect to $t$ is $\frac{d}{d t} [- 1] + \frac{d}{d t} [- 13 t]$ .

$\frac{d}{d t} [- 1] + \frac{d}{d t} [- 13 t]$

Step 1.1.2

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

$0 + \frac{d}{d t} [- 13 t]$

Step 1.2

Evaluate $\frac{d}{d t} [- 13 t]$ .

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Step 1.2.1

Since $- 13$ is constant with respect to $t$ , the derivative of $- 13 t$ with respect to $t$ is $- 13 \frac{d}{d t} [t]$ .

$0 - 13 \frac{d}{d t} [t]$

Step 1.2.2

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

$0 - 13 \cdot 1$

Step 1.2.3

Multiply $- 13$ by $1$ .

$0 - 13$

Step 1.3

Subtract $13$ from $0$ .

$- 13$

Step 2

Replace the variable $t$ with $8$ in the expression.

$s' (8) = - 13$