Calculus Examples

$y = x^{2} \ln (4 x)$

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

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

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 $g (x) = 4 x$ .

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To apply the Chain Rule, set $u$ as $4 x$ .

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

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

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

Replace all occurrences of $u$ with $4 x$ .

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

Step 3

Differentiate.

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

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

Cancel the common factor of $x^{2}$ and $x$ .

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Factor $x$ out of $x^{2}$ .

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

Cancel the common factors.

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Factor $x$ out of $4 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Simplify terms.

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Combine $4$ and $\frac{x}{4}$ .

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

Cancel the common factor of $4$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Multiply $x$ by $1$ .

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

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

$x + \ln (4 x) (2 x)$

Reorder terms.

$2 x \ln (4 x) + x$