Calculus Examples

$f (t) = - 5 t^{2} + 48 t + 25$

Step 1

This derivative could not be completed using the chain rule. Mathway will use another method.

Step 2

By the Sum Rule, the derivative of $- 5 t^{2} + 48 t + 25$ with respect to $t$ is $\frac{d}{d t} [- 5 t^{2}] + \frac{d}{d t} [48 t] + \frac{d}{d t} [25]$ .

$\frac{d}{d t} [- 5 t^{2}] + \frac{d}{d t} [48 t] + \frac{d}{d t} [25]$

Step 3

Evaluate $\frac{d}{d t} [- 5 t^{2}]$ .

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Step 3.1

Since $- 5$ is constant with respect to $t$ , the derivative of $- 5 t^{2}$ with respect to $t$ is $- 5 \frac{d}{d t} [t^{2}]$ .

$- 5 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [48 t] + \frac{d}{d t} [25]$

Step 3.2

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

$- 5 (2 t) + \frac{d}{d t} [48 t] + \frac{d}{d t} [25]$

Step 3.3

Multiply $2$ by $- 5$ .

$- 10 t + \frac{d}{d t} [48 t] + \frac{d}{d t} [25]$

Step 4

Evaluate $\frac{d}{d t} [48 t]$ .

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Step 4.1

Since $48$ is constant with respect to $t$ , the derivative of $48 t$ with respect to $t$ is $48 \frac{d}{d t} [t]$ .

$- 10 t + 48 \frac{d}{d t} [t] + \frac{d}{d t} [25]$

Step 4.2

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

$- 10 t + 48 \cdot 1 + \frac{d}{d t} [25]$

Step 4.3

Multiply $48$ by $1$ .

$- 10 t + 48 + \frac{d}{d t} [25]$

Step 5

Differentiate using the Constant Rule.

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Step 5.1

Since $25$ is constant with respect to $t$ , the derivative of $25$ with respect to $t$ is $0$ .

$- 10 t + 48 + 0$

Step 5.2

Add $- 10 t + 48$ and $0$ .

$- 10 t + 48$