Calculus Examples

$(a t^{2} + b t + c)$

Step 1

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

$\frac{d}{d t} [a t^{2}] + \frac{d}{d t} [b t] + \frac{d}{d t} [c]$

Step 2

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

$a \frac{d}{d t} [t^{2}] + \frac{d}{d t} [b t] + \frac{d}{d t} [c]$

Step 3

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

$a (2 t) + \frac{d}{d t} [b t] + \frac{d}{d t} [c]$

Step 4

Move $2$ to the left of $a$ .

$2 \cdot a t + \frac{d}{d t} [b t] + \frac{d}{d t} [c]$

Step 5

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

$2 a t + b \frac{d}{d t} [t] + \frac{d}{d t} [c]$

Step 6

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

$2 a t + b \cdot 1 + \frac{d}{d t} [c]$

Step 7

Multiply $b$ by $1$ .

$2 a t + b + \frac{d}{d t} [c]$

Step 8

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

$2 a t + b + 0$

Step 9

Simplify the expression.

Tap for more steps...

Step 9.1

Add $2 a t + b$ and $0$ .

$2 a t + b$

Step 9.2

Reorder terms.

$b + 2 a t$