Calculus Examples

$y = x \ln (x)$

Step 1

Find the first derivative.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Differentiate using the Power Rule.

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Step 1.3.1

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

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

Step 1.3.2

Cancel the common factor of $x$ .

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Step 1.3.2.1

Cancel the common factor.

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

Step 1.3.2.2

Rewrite the expression.

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

Step 1.3.3

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

$1 + \ln (x) \cdot 1$

Step 1.3.4

Multiply $\ln (x)$ by $1$ .

$f' (x) = 1 + \ln (x)$

Step 2

Find the second derivative.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $1 + \ln (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\ln (x)]$ .

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

Step 2.1.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\ln (x)]$

Step 2.2

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

$0 + \frac{1}{x}$

Step 2.3

Add $0$ and $\frac{1}{x}$ .

$f'' (x) = \frac{1}{x}$

Step 3

Find the third derivative.

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Step 3.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$\frac{d}{d x} [x^{- 1}]$

Step 3.2

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

$- x^{- 2}$

Step 3.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f''' (x) = - \frac{1}{x^{2}}$

Step 4

Find the fourth derivative.

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Step 4.1

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 $f (x) = - 1$ and $g (x) = \frac{1}{x^{2}}$ .

$- \frac{d}{d x} [\frac{1}{x^{2}}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 4.2

Differentiate.

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Step 4.2.1

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

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

Step 4.2.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 4.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{d}{d x} [x^{2 \cdot - 1}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 4.2.2.2

Multiply $2$ by $- 1$ .

$- \frac{d}{d x} [x^{- 2}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 4.2.3

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

$- (- 2 x^{- 3}) + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 4.2.4

Multiply $- 2$ by $- 1$ .

$2 x^{- 3} + \frac{1}{x^{2}} \frac{d}{d x} [- 1]$

Step 4.2.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 x^{- 3} + \frac{1}{x^{2}} \cdot 0$

Step 4.2.6

Simplify the expression.

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Step 4.2.6.1

Multiply $\frac{1}{x^{2}}$ by $0$ .

$2 x^{- 3} + 0$

Step 4.2.6.2

Add $2 x^{- 3}$ and $0$ .

$2 x^{- 3}$

Step 4.3

Simplify.

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Step 4.3.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 \frac{1}{x^{3}}$

Step 4.3.2

Combine $2$ and $\frac{1}{x^{3}}$ .

$f^{4} (x) = \frac{2}{x^{3}}$