Calculus Examples

$g (t) = t^{4} + t^{3} + t^{2} + 1$

Step 1

By the Sum Rule, the derivative of $t^{4} + t^{3} + t^{2} + 1$ with respect to $t$ is $\frac{d}{d t} [t^{4}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$ .

$\frac{d}{d t} [t^{4}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 4$ .

$4 t^{3} + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 3

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

$4 t^{3} + 3 t^{2} + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 4

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

$4 t^{3} + 3 t^{2} + 2 t + \frac{d}{d t} [1]$

Step 5

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

$4 t^{3} + 3 t^{2} + 2 t + 0$

Step 6

Add $4 t^{3} + 3 t^{2} + 2 t$ and $0$ .

$4 t^{3} + 3 t^{2} + 2 t$