Calculus Examples

$g (t) = 4 t - \frac{1}{8} + 2 \ln (t) + \ln (5)$

Step 1

By the Sum Rule, the derivative of $4 t - \frac{1}{8} + 2 \ln (t) + \ln (5)$ with respect to $t$ is $\frac{d}{d t} [4 t] + \frac{d}{d t} [- \frac{1}{8}] + \frac{d}{d t} [2 \ln (t)] + \frac{d}{d t} [\ln (5)]$ .

$\frac{d}{d t} [4 t] + \frac{d}{d t} [- \frac{1}{8}] + \frac{d}{d t} [2 \ln (t)] + \frac{d}{d t} [\ln (5)]$

Step 2

Evaluate $\frac{d}{d t} [4 t]$ .

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Step 2.1

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

$4 \frac{d}{d t} [t] + \frac{d}{d t} [- \frac{1}{8}] + \frac{d}{d t} [2 \ln (t)] + \frac{d}{d t} [\ln (5)]$

Step 2.2

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

$4 \cdot 1 + \frac{d}{d t} [- \frac{1}{8}] + \frac{d}{d t} [2 \ln (t)] + \frac{d}{d t} [\ln (5)]$

Step 2.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d t} [- \frac{1}{8}] + \frac{d}{d t} [2 \ln (t)] + \frac{d}{d t} [\ln (5)]$

Step 3

Since $- \frac{1}{8}$ is constant with respect to $t$ , the derivative of $- \frac{1}{8}$ with respect to $t$ is $0$ .

$4 + 0 + \frac{d}{d t} [2 \ln (t)] + \frac{d}{d t} [\ln (5)]$

Step 4

Evaluate $\frac{d}{d t} [2 \ln (t)]$ .

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Step 4.1

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

$4 + 0 + 2 \frac{d}{d t} [\ln (t)] + \frac{d}{d t} [\ln (5)]$

Step 4.2

The derivative of $\ln (t)$ with respect to $t$ is $\frac{1}{t}$ .

$4 + 0 + 2 \frac{1}{t} + \frac{d}{d t} [\ln (5)]$

Step 4.3

Combine $2$ and $\frac{1}{t}$ .

$4 + 0 + \frac{2}{t} + \frac{d}{d t} [\ln (5)]$

Step 5

Since $\ln (5)$ is constant with respect to $t$ , the derivative of $\ln (5)$ with respect to $t$ is $0$ .

$4 + 0 + \frac{2}{t} + 0$

Step 6

Simplify.

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Step 6.1

Combine terms.

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Step 6.1.1

Add $4$ and $0$ .

$4 + \frac{2}{t} + 0$

Step 6.1.2

Add $4 + \frac{2}{t}$ and $0$ .

$4 + \frac{2}{t}$

Step 6.2

Reorder terms.

$\frac{2}{t} + 4$