Calculus Examples

$x \ln (8 x)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (8 x)$ .

$x \frac{d}{d x} [\ln (8 x)] + \ln (8 x) \frac{d}{d x} [x]$

Step 2

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 $g (x) = 8 x$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $8 x$ .

$x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [8 x]) + \ln (8 x) \frac{d}{d x} [x]$

Step 2.2

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

$x (\frac{1}{u} \frac{d}{d x} [8 x]) + \ln (8 x) \frac{d}{d x} [x]$

Step 2.3

Replace all occurrences of $u$ with $8 x$ .

$x (\frac{1}{8 x} \frac{d}{d x} [8 x]) + \ln (8 x) \frac{d}{d x} [x]$

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{8 x}$ and $x$ .

$\frac{x}{8 x} \frac{d}{d x} [8 x] + \ln (8 x) \frac{d}{d x} [x]$

Step 3.2

Cancel the common factor of $x$ .

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Step 3.2.1

Cancel the common factor.

$\frac{x}{8 x} \frac{d}{d x} [8 x] + \ln (8 x) \frac{d}{d x} [x]$

Step 3.2.2

Rewrite the expression.

$\frac{1}{8} \frac{d}{d x} [8 x] + \ln (8 x) \frac{d}{d x} [x]$

Step 3.3

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$\frac{1}{8} (8 \frac{d}{d x} [x]) + \ln (8 x) \frac{d}{d x} [x]$

Step 3.4

Simplify terms.

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Step 3.4.1

Combine $8$ and $\frac{1}{8}$ .

$\frac{8}{8} \frac{d}{d x} [x] + \ln (8 x) \frac{d}{d x} [x]$

Step 3.4.2

Cancel the common factor of $8$ .

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Step 3.4.2.1

Cancel the common factor.

$\frac{8}{8} \frac{d}{d x} [x] + \ln (8 x) \frac{d}{d x} [x]$

Step 3.4.2.2

Rewrite the expression.

$1 \frac{d}{d x} [x] + \ln (8 x) \frac{d}{d x} [x]$

Step 3.4.3

Multiply $\frac{d}{d x} [x]$ by $1$ .

$\frac{d}{d x} [x] + \ln (8 x) \frac{d}{d x} [x]$

Step 3.5

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

$1 + \ln (8 x) \frac{d}{d x} [x]$

Step 3.6

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

$1 + \ln (8 x) \cdot 1$

Step 3.7

Multiply $\ln (8 x)$ by $1$ .

$1 + \ln (8 x)$