Precalculus Examples

$h (t) = 12 t^{2} - 6 t + 50$

Step 1

By the Sum Rule, the derivative of $12 t^{2} - 6 t + 50$ with respect to $t$ is $\frac{d}{d t} [12 t^{2}] + \frac{d}{d t} [- 6 t] + \frac{d}{d t} [50]$ .

$\frac{d}{d t} [12 t^{2}] + \frac{d}{d t} [- 6 t] + \frac{d}{d t} [50]$

Step 2

Evaluate $\frac{d}{d t} [12 t^{2}]$ .

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Step 2.1

Since $12$ is constant with respect to $t$ , the derivative of $12 t^{2}$ with respect to $t$ is $12 \frac{d}{d t} [t^{2}]$ .

$12 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 6 t] + \frac{d}{d t} [50]$

Step 2.2

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

$12 (2 t) + \frac{d}{d t} [- 6 t] + \frac{d}{d t} [50]$

Step 2.3

Multiply $2$ by $12$ .

$24 t + \frac{d}{d t} [- 6 t] + \frac{d}{d t} [50]$

Step 3

Evaluate $\frac{d}{d t} [- 6 t]$ .

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Step 3.1

Since $- 6$ is constant with respect to $t$ , the derivative of $- 6 t$ with respect to $t$ is $- 6 \frac{d}{d t} [t]$ .

$24 t - 6 \frac{d}{d t} [t] + \frac{d}{d t} [50]$

Step 3.2

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

$24 t - 6 \cdot 1 + \frac{d}{d t} [50]$

Step 3.3

Multiply $- 6$ by $1$ .

$24 t - 6 + \frac{d}{d t} [50]$

Step 4

Differentiate using the Constant Rule.

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Step 4.1

Since $50$ is constant with respect to $t$ , the derivative of $50$ with respect to $t$ is $0$ .

$24 t - 6 + 0$

Step 4.2

Add $24 t - 6$ and $0$ .

$24 t - 6$