Calculus Examples

$\ln (t + 1) - 0.5 t$

Step 1

By the Sum Rule, the derivative of $\ln (t + 1) - 0.5 t$ with respect to $t$ is $\frac{d}{d t} [\ln (t + 1)] + \frac{d}{d t} [- 0.5 t]$ .

$\frac{d}{d t} [\ln (t + 1)] + \frac{d}{d t} [- 0.5 t]$

Step 2

Evaluate $\frac{d}{d t} [\ln (t + 1)]$ .

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \ln (t)$ and $g (t) = t + 1$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $t + 1$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d t} [t + 1] + \frac{d}{d t} [- 0.5 t]$

Step 2.1.2

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

$\frac{1}{u} \frac{d}{d t} [t + 1] + \frac{d}{d t} [- 0.5 t]$

Step 2.1.3

Replace all occurrences of $u$ with $t + 1$ .

$\frac{1}{t + 1} \frac{d}{d t} [t + 1] + \frac{d}{d t} [- 0.5 t]$

Step 2.2

By the Sum Rule, the derivative of $t + 1$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [1]$ .

$\frac{1}{t + 1} (\frac{d}{d t} [t] + \frac{d}{d t} [1]) + \frac{d}{d t} [- 0.5 t]$

Step 2.3

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

$\frac{1}{t + 1} (1 + \frac{d}{d t} [1]) + \frac{d}{d t} [- 0.5 t]$

Step 2.4

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$\frac{1}{t + 1} (1 + 0) + \frac{d}{d t} [- 0.5 t]$

Step 2.5

Add $1$ and $0$ .

$\frac{1}{t + 1} \cdot 1 + \frac{d}{d t} [- 0.5 t]$

Step 2.6

Multiply $\frac{1}{t + 1}$ by $1$ .

$\frac{1}{t + 1} + \frac{d}{d t} [- 0.5 t]$

Step 3

Evaluate $\frac{d}{d t} [- 0.5 t]$ .

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Step 3.1

Since $- 0.5$ is constant with respect to $t$ , the derivative of $- 0.5 t$ with respect to $t$ is $- 0.5 \frac{d}{d t} [t]$ .

$\frac{1}{t + 1} - 0.5 \frac{d}{d t} [t]$

Step 3.2

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

$\frac{1}{t + 1} - 0.5 \cdot 1$

Step 3.3

Multiply $- 0.5$ by $1$ .

$\frac{1}{t + 1} - 0.5$

Step 4

Combine terms.

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Step 4.1

To write $- 0.5$ as a fraction with a common denominator, multiply by $\frac{t + 1}{t + 1}$ .

$\frac{1}{t + 1} - 0.5 \cdot \frac{t + 1}{t + 1}$

Step 4.2

Combine $- 0.5$ and $\frac{t + 1}{t + 1}$ .

$\frac{1}{t + 1} + \frac{- 0.5 (t + 1)}{t + 1}$

Step 4.3

Combine the numerators over the common denominator.

$\frac{1 - 0.5 (t + 1)}{t + 1}$