Calculus Examples

$y = x^{\cot (x)}$

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

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

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

Step 1.2

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

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

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 $g (x) = \cot (x) \ln (x)$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $\cot (x) \ln (x)$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $\cot (x) \ln (x)$ .

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

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

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

Step 4

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

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

Step 5

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

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

Step 6

The derivative of $\cot (x)$ with respect to $x$ is $- \csc^{2} (x)$ .

$e^{\cot (x) \ln (x)} (\frac{\cot (x)}{x} + \ln (x) (- \csc^{2} (x)))$

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

$e^{\cot (x) \ln (x)} \frac{\cot (x)}{x} + e^{\cot (x) \ln (x)} (\ln (x) (- \csc^{2} (x)))$

Step 7.2

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

$\frac{e^{\cot (x) \ln (x)} \cot (x)}{x} + e^{\cot (x) \ln (x)} \ln (x) (- \csc^{2} (x))$

Step 7.3

Reorder terms.

$- e^{\cot (x) \ln (x)} \csc^{2} (x) \ln (x) + \frac{e^{\cot (x) \ln (x)} \cot (x)}{x}$