Calculus Examples

$f (x) = x^{x^{4}}$

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

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

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

Step 1.2

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

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

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $x^{4} \ln (x)$ .

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

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

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

Step 4

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

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

Step 5

Differentiate using the Power Rule.

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Step 5.1

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

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

Step 5.2

Cancel the common factor of $x^{4}$ and $x$ .

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Step 5.2.1

Factor $x$ out of $x^{4}$ .

$e^{x^{4} \ln (x)} (\frac{x \cdot x^{3}}{x} + \ln (x) \frac{d}{d x} [x^{4}])$

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Raise $x$ to the power of $1$ .

$e^{x^{4} \ln (x)} (\frac{x \cdot x^{3}}{x^{1}} + \ln (x) \frac{d}{d x} [x^{4}])$

Step 5.2.2.2

Factor $x$ out of $x^{1}$ .

$e^{x^{4} \ln (x)} (\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{4}])$

Step 5.2.2.3

Cancel the common factor.

$e^{x^{4} \ln (x)} (\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{4}])$

Step 5.2.2.4

Rewrite the expression.

$e^{x^{4} \ln (x)} (\frac{x^{3}}{1} + \ln (x) \frac{d}{d x} [x^{4}])$

Step 5.2.2.5

Divide $x^{3}$ by $1$ .

$e^{x^{4} \ln (x)} (x^{3} + \ln (x) \frac{d}{d x} [x^{4}])$

Step 5.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$e^{x^{4} \ln (x)} (x^{3} + \ln (x) (4 x^{3}))$

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

$e^{x^{4} \ln (x)} x^{3} + e^{x^{4} \ln (x)} (\ln (x) (4 x^{3}))$

Step 6.2

Reorder terms.

$e^{x^{4} \ln (x)} x^{3} + 4 e^{x^{4} \ln (x)} x^{3} \ln (x)$