Calculus Examples

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

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

Step 1

By the Sum Rule, the derivative of

t^{4} + t^{3} + t^{2} + 1

$t^{4} + t^{3} + t^{2} + 1$ with respect to

t

$t$ is

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

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

\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}]

$\frac{d}{d t} [t^{n}]$ is

n t^{n - 1}

$n t^{n - 1}$ where

n = 4

$n = 4$ .

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

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

Step 3

Differentiate using the Power Rule which states that

\frac{d}{d t} [t^{n}]

$\frac{d}{d t} [t^{n}]$ is

n t^{n - 1}

$n t^{n - 1}$ where

n = 3

$n = 3$ .

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

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

Step 4

Differentiate using the Power Rule which states that

\frac{d}{d t} [t^{n}]

$\frac{d}{d t} [t^{n}]$ is

n t^{n - 1}

$n t^{n - 1}$ where

n = 2

$n = 2$ .

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

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

Step 5

Since

1

$1$ is constant with respect to

t

$t$ , the derivative of

1

$1$ with respect to

t

$t$ is

0

$0$ .

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

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

Step 6

Add

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

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

0

$0$ .

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

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