Calculus Examples

$h (t) = (t^{9} - 10 t^{4} + 4 t + 3)$

Step 1

By the Sum Rule, the derivative of $t^{9} - 10 t^{4} + 4 t + 3$ with respect to $t$ is $\frac{d}{d t} [t^{9}] + \frac{d}{d t} [- 10 t^{4}] + \frac{d}{d t} [4 t] + \frac{d}{d t} [3]$ .

$\frac{d}{d t} [t^{9}] + \frac{d}{d t} [- 10 t^{4}] + \frac{d}{d t} [4 t] + \frac{d}{d t} [3]$

Step 2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 9$ .

$9 t^{8} + \frac{d}{d t} [- 10 t^{4}] + \frac{d}{d t} [4 t] + \frac{d}{d t} [3]$

Step 3

Since $- 10$ is constant with respect to $t$ , the derivative of $- 10 t^{4}$ with respect to $t$ is $- 10 \frac{d}{d t} [t^{4}]$ .

$9 t^{8} - 10 \frac{d}{d t} [t^{4}] + \frac{d}{d t} [4 t] + \frac{d}{d t} [3]$

Step 4

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

$9 t^{8} - 10 (4 t^{3}) + \frac{d}{d t} [4 t] + \frac{d}{d t} [3]$

Step 5

Multiply $4$ by $- 10$ .

$9 t^{8} - 40 t^{3} + \frac{d}{d t} [4 t] + \frac{d}{d t} [3]$

Step 6

Since $4$ is constant with respect to $t$ , the derivative of $4 t$ with respect to $t$ is $4 \frac{d}{d t} [t]$ .

$9 t^{8} - 40 t^{3} + 4 \frac{d}{d t} [t] + \frac{d}{d t} [3]$

Step 7

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

$9 t^{8} - 40 t^{3} + 4 \cdot 1 + \frac{d}{d t} [3]$

Step 8

Multiply $4$ by $1$ .

$9 t^{8} - 40 t^{3} + 4 + \frac{d}{d t} [3]$

Step 9

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$9 t^{8} - 40 t^{3} + 4 + 0$

Step 10

Add $9 t^{8} - 40 t^{3} + 4$ and $0$ .

$9 t^{8} - 40 t^{3} + 4$