Calculus Examples

$f (x) = 20 x + \ln (20 x)$

Step 1

By the Sum Rule, the derivative of $20 x + \ln (20 x)$ with respect to $x$ is $\frac{d}{d x} [20 x] + \frac{d}{d x} [\ln (20 x)]$ .

$\frac{d}{d x} [20 x] + \frac{d}{d x} [\ln (20 x)]$

Step 2

Evaluate $\frac{d}{d x} [20 x]$ .

Tap for more steps...

Step 2.1

Since $20$ is constant with respect to $x$ , the derivative of $20 x$ with respect to $x$ is $20 \frac{d}{d x} [x]$ .

$20 \frac{d}{d x} [x] + \frac{d}{d x} [\ln (20 x)]$

Step 2.2

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

$20 \cdot 1 + \frac{d}{d x} [\ln (20 x)]$

Step 2.3

Multiply $20$ by $1$ .

$20 + \frac{d}{d x} [\ln (20 x)]$

Step 3

Evaluate $\frac{d}{d x} [\ln (20 x)]$ .

Tap for more steps...

Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 20 x$ .

Tap for more steps...

Step 3.1.1

To apply the Chain Rule, set $u$ as $20 x$ .

$20 + \frac{d}{d u} [\ln (u)] \frac{d}{d x} [20 x]$

Step 3.1.2

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

$20 + \frac{1}{u} \frac{d}{d x} [20 x]$

Step 3.1.3

Replace all occurrences of $u$ with $20 x$ .

$20 + \frac{1}{20 x} \frac{d}{d x} [20 x]$

Step 3.2

Since $20$ is constant with respect to $x$ , the derivative of $20 x$ with respect to $x$ is $20 \frac{d}{d x} [x]$ .

$20 + \frac{1}{20 x} (20 \frac{d}{d x} [x])$

Step 3.3

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

$20 + \frac{1}{20 x} (20 \cdot 1)$

Step 3.4

Multiply $20$ by $1$ .

$20 + \frac{1}{20 x} \cdot 20$

Step 3.5

Combine $\frac{1}{20 x}$ and $20$ .

$20 + \frac{20}{20 x}$

Step 3.6

Cancel the common factor of $20$ .

Tap for more steps...

Step 3.6.1

Cancel the common factor.

$20 + \frac{20}{20 x}$

Step 3.6.2

Rewrite the expression.

$20 + \frac{1}{x}$

Step 4

Reorder terms.

$\frac{1}{x} + 20$