Calculus Examples

$x \ln (x^{2} + 4 x)$

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 $g (x) = \ln (x^{2} + 4 x)$ .

$x \frac{d}{d x} [\ln (x^{2} + 4 x)] + \ln (x^{2} + 4 x) \frac{d}{d x} [x]$

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 $g (x) = x^{2} + 4 x$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $x^{2} + 4 x$ .

$x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{2} + 4 x]) + \ln (x^{2} + 4 x) \frac{d}{d x} [x]$

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $x^{2} + 4 x$ .

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

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{x^{2} + 4 x}$ and $x$ .

$\frac{x}{x^{2} + 4 x} \frac{d}{d x} [x^{2} + 4 x] + \ln (x^{2} + 4 x) \frac{d}{d x} [x]$

Step 3.2

By the Sum Rule, the derivative of $x^{2} + 4 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x]$ .

$\frac{x}{x^{2} + 4 x} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x]) + \ln (x^{2} + 4 x) \frac{d}{d x} [x]$

Step 3.3

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

$\frac{x}{x^{2} + 4 x} (2 x + \frac{d}{d x} [4 x]) + \ln (x^{2} + 4 x) \frac{d}{d x} [x]$

Step 3.4

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

$\frac{x}{x^{2} + 4 x} (2 x + 4 \frac{d}{d x} [x]) + \ln (x^{2} + 4 x) \frac{d}{d x} [x]$

Step 3.5

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

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

Step 3.6

Multiply $4$ by $1$ .

$\frac{x}{x^{2} + 4 x} (2 x + 4) + \ln (x^{2} + 4 x) \frac{d}{d x} [x]$

Step 3.7

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

$\frac{x}{x^{2} + 4 x} (2 x + 4) + \ln (x^{2} + 4 x) \cdot 1$

Step 3.8

Simplify the expression.

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Step 3.8.1

Multiply $\ln (x^{2} + 4 x)$ by $1$ .

$\frac{x}{x^{2} + 4 x} (2 x + 4) + \ln (x^{2} + 4 x)$

Step 3.8.2

Reorder terms.

$(2 x + 4) \frac{x}{x^{2} + 4 x} + \ln (x^{2} + 4 x)$