Calculus Examples

$g (t) = \frac{t}{3 - t} + 5 t^{3}$

Step 1

By the Sum Rule, the derivative of $\frac{t}{3 - t} + 5 t^{3}$ with respect to $t$ is $\frac{d}{d t} [\frac{t}{3 - t}] + \frac{d}{d t} [5 t^{3}]$ .

$\frac{d}{d t} [\frac{t}{3 - t}] + \frac{d}{d t} [5 t^{3}]$

Step 2

Evaluate $\frac{d}{d t} [\frac{t}{3 - t}]$ .

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Step 2.1

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t$ and $g (t) = 3 - t$ .

$\frac{(3 - t) \frac{d}{d t} [t] - t \frac{d}{d t} [3 - t]}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.2

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

$\frac{(3 - t) \cdot 1 - t \frac{d}{d t} [3 - t]}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.3

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

$\frac{(3 - t) \cdot 1 - t (\frac{d}{d t} [3] + \frac{d}{d t} [- t])}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.4

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

$\frac{(3 - t) \cdot 1 - t (0 + \frac{d}{d t} [- t])}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.5

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$\frac{(3 - t) \cdot 1 - t (0 - \frac{d}{d t} [t])}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.6

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

$\frac{(3 - t) \cdot 1 - t (0 - 1 \cdot 1)}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.7

Multiply $3 - t$ by $1$ .

$\frac{3 - t - t (0 - 1 \cdot 1)}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.8

Multiply $- 1$ by $1$ .

$\frac{3 - t - t (0 - 1)}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.9

Subtract $1$ from $0$ .

$\frac{3 - t - t \cdot - 1}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.10

Multiply $- 1$ by $- 1$ .

$\frac{3 - t + 1 t}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.11

Multiply $t$ by $1$ .

$\frac{3 - t + t}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.12

Add $- t$ and $t$ .

$\frac{3 + 0}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 2.13

Add $3$ and $0$ .

$\frac{3}{{(3 - t)}^{2}} + \frac{d}{d t} [5 t^{3}]$

Step 3

Evaluate $\frac{d}{d t} [5 t^{3}]$ .

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Step 3.1

Since $5$ is constant with respect to $t$ , the derivative of $5 t^{3}$ with respect to $t$ is $5 \frac{d}{d t} [t^{3}]$ .

$\frac{3}{{(3 - t)}^{2}} + 5 \frac{d}{d t} [t^{3}]$

Step 3.2

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

$\frac{3}{{(3 - t)}^{2}} + 5 (3 t^{2})$

Step 3.3

Multiply $3$ by $5$ .

$\frac{3}{{(3 - t)}^{2}} + 15 t^{2}$

Step 4

Reorder terms.

$15 t^{2} + \frac{3}{{(3 - t)}^{2}}$