Calculus Examples

$s = 72 + 64 t - 16 t^{2}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (s) = \frac{d}{d t} (72 + 64 t - 16 t^{2})$

Step 2

The derivative of $s$ with respect to $t$ is $s'$ .

$s'$

Step 3

Differentiate the right side of the equation.

Tap for more steps...

Step 3.1

Differentiate.

Tap for more steps...

Step 3.1.1

By the Sum Rule, the derivative of $72 + 64 t - 16 t^{2}$ with respect to $t$ is $\frac{d}{d t} [72] + \frac{d}{d t} [64 t] + \frac{d}{d t} [- 16 t^{2}]$ .

$\frac{d}{d t} [72] + \frac{d}{d t} [64 t] + \frac{d}{d t} [- 16 t^{2}]$

Step 3.1.2

Since $72$ is constant with respect to $t$ , the derivative of $72$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [64 t] + \frac{d}{d t} [- 16 t^{2}]$

Step 3.2

Evaluate $\frac{d}{d t} [64 t]$ .

Tap for more steps...

Step 3.2.1

Since $64$ is constant with respect to $t$ , the derivative of $64 t$ with respect to $t$ is $64 \frac{d}{d t} [t]$ .

$0 + 64 \frac{d}{d t} [t] + \frac{d}{d t} [- 16 t^{2}]$

Step 3.2.2

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

$0 + 64 \cdot 1 + \frac{d}{d t} [- 16 t^{2}]$

Step 3.2.3

Multiply $64$ by $1$ .

$0 + 64 + \frac{d}{d t} [- 16 t^{2}]$

Step 3.3

Evaluate $\frac{d}{d t} [- 16 t^{2}]$ .

Tap for more steps...

Step 3.3.1

Since $- 16$ is constant with respect to $t$ , the derivative of $- 16 t^{2}$ with respect to $t$ is $- 16 \frac{d}{d t} [t^{2}]$ .

$0 + 64 - 16 \frac{d}{d t} [t^{2}]$

Step 3.3.2

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

$0 + 64 - 16 (2 t)$

Step 3.3.3

Multiply $2$ by $- 16$ .

$0 + 64 - 32 t$

Step 3.4

Simplify.

Tap for more steps...

Step 3.4.1

Add $0$ and $64$ .

$64 - 32 t$

Step 3.4.2

Reorder terms.

$- 32 t + 64$

Step 4

Reform the equation by setting the left side equal to the right side.

$s' = - 32 t + 64$

Step 5

Replace $s'$ with $\frac{d s}{d t}$ .

$\frac{d s}{d t} = - 32 t + 64$