Calculus Examples

$\frac{1}{4} t^{2} - \ln (t + 1)$

Step 1

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

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

Step 2

Evaluate $\frac{d}{d t} [\frac{1}{4} t^{2}]$ .

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Step 2.1

Since $\frac{1}{4}$ is constant with respect to $t$ , the derivative of $\frac{1}{4} t^{2}$ with respect to $t$ is $\frac{1}{4} \frac{d}{d t} [t^{2}]$ .

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

Step 2.2

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

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

Step 2.3

Combine $2$ and $\frac{1}{4}$ .

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

Step 2.4

Combine $\frac{2}{4}$ and $t$ .

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

Step 2.5

Cancel the common factor of $2$ and $4$ .

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Step 2.5.1

Factor $2$ out of $2 t$ .

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

Step 2.5.2

Cancel the common factors.

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Step 2.5.2.1

Factor $2$ out of $4$ .

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

Step 2.5.2.2

Cancel the common factor.

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

Step 2.5.2.3

Rewrite the expression.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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 3.2.1

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

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

Step 3.2.2

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

$\frac{t}{2} - (\frac{1}{u} \frac{d}{d t} [t + 1])$

Step 3.2.3

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

$\frac{t}{2} - (\frac{1}{t + 1} (1 + 0))$

Step 3.6

Add $1$ and $0$ .

$\frac{t}{2} - (\frac{1}{t + 1} \cdot 1)$

Step 3.7

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

$\frac{t}{2} - \frac{1}{t + 1}$

Step 4

Combine terms.

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Step 4.1

To write $\frac{t}{2}$ as a fraction with a common denominator, multiply by $\frac{t + 1}{t + 1}$ .

$\frac{t}{2} \cdot \frac{t + 1}{t + 1} - \frac{1}{t + 1}$

Step 4.2

To write $- \frac{1}{t + 1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{t}{2} \cdot \frac{t + 1}{t + 1} - \frac{1}{t + 1} \cdot \frac{2}{2}$

Step 4.3

Write each expression with a common denominator of $2 (t + 1)$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.3.1

Multiply $\frac{t}{2}$ by $\frac{t + 1}{t + 1}$ .

$\frac{t (t + 1)}{2 (t + 1)} - \frac{1}{t + 1} \cdot \frac{2}{2}$

Step 4.3.2

Multiply $\frac{1}{t + 1}$ by $\frac{2}{2}$ .

$\frac{t (t + 1)}{2 (t + 1)} - \frac{2}{(t + 1) \cdot 2}$

Step 4.3.3

Reorder the factors of $(t + 1) \cdot 2$ .

$\frac{t (t + 1)}{2 (t + 1)} - \frac{2}{2 (t + 1)}$

Step 4.4

Combine the numerators over the common denominator.

$\frac{t (t + 1) - 2}{2 (t + 1)}$