Calculus Examples

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

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

Rewrite $x^{\ln (4 x)}$ as $e^{\ln (x^{\ln (4 x)})}$ .

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

Step 1.2

Expand $\ln (x^{\ln (4 x)})$ by moving $\ln (4 x)$ outside the logarithm.

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = \ln (4 x) \ln (x)$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $\ln (4 x) \ln (x)$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [\ln (4 x) \ln (x)]$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [\ln (4 x) \ln (x)]$

Step 2.3

Replace all occurrences of $u_{1}$ with $\ln (4 x) \ln (x)$ .

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

Step 3

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 $f (x) = \ln (4 x)$ and $g (x) = \ln (x)$ .

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

Step 4

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

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

Step 5

Combine $\ln (4 x)$ and $\frac{1}{x}$ .

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

Step 6

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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Step 6.1

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

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

Step 6.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

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

Step 6.3

Replace all occurrences of $u_{2}$ with $4 x$ .

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

Step 7

Differentiate.

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Step 7.1

Combine $\frac{1}{4 x}$ and $\ln (x)$ .

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

Step 7.2

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

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

Step 7.3

Simplify terms.

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Step 7.3.1

Combine $4$ and $\frac{\ln (x)}{4 x}$ .

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

Step 7.3.2

Cancel the common factor of $4$ .

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Step 7.3.2.1

Cancel the common factor.

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

Step 7.3.2.2

Rewrite the expression.

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

Step 7.4

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

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

Step 7.5

Multiply $\frac{\ln (x)}{x}$ by $1$ .

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

Step 8

Simplify.

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Step 8.1

Apply the distributive property.

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

Step 8.2

Combine terms.

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Step 8.2.1

Combine $e^{\ln (4 x) \ln (x)}$ and $\frac{\ln (4 x)}{x}$ .

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

Step 8.2.2

Combine $e^{\ln (4 x) \ln (x)}$ and $\frac{\ln (x)}{x}$ .

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