Find the 4th Derivative f(x) = natural log of x

$f (x) = \ln (x)$

Step 1

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

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

Step 2

Find the second derivative.

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

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

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

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

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

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

Step 3

Find the third derivative.

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

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

Differentiate.

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

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

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

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

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

Multiply $2$ by $- 1$ .

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

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

$f''' (x) = - (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

Multiply $- 2$ by $- 1$ .

$f''' (x) = 2 x^{- 3} + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

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

$f''' (x) = 2 x^{- 3} + \frac{1}{x^{2}} \cdot 0$

Simplify the expression.

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

$f''' (x) = 2 x^{- 3} + 0$

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

$f''' (x) = 2 x^{- 3}$

Simplify.

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

$f''' (x) = 2 (\frac{1}{x^{3}})$

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

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

Step 4

Find the fourth derivative.

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2}{x^{3}}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{1}{x^{3}}]$ .

$f^{4} (x) = 2 \frac{d}{d x} (\frac{1}{x^{3}})$

Apply basic rules of exponents.

Rewrite $\frac{1}{x^{3}}$ as ${(x^{3})}^{- 1}$ .

$f^{4} (x) = 2 \frac{d}{d x} ({(x^{3})}^{- 1})$

Multiply the exponents in ${(x^{3})}^{- 1}$ .

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

$f^{4} (x) = 2 \frac{d}{d x} (x^{3 \cdot - 1})$

Multiply $3$ by $- 1$ .

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

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 3$ .

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

Multiply $- 3$ by $2$ .

$f^{4} (x) = - 6 x^{- 4}$

Simplify.

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

$f^{4} (x) = - 6 \frac{1}{x^{4}}$

Combine terms.

Combine $- 6$ and $\frac{1}{x^{4}}$ .

$f^{4} (x) = \frac{- 6}{x^{4}}$

Move the negative in front of the fraction.

$f^{4} (x) = - \frac{6}{x^{4}}$

Step 5

The fourth derivative of $f (x)$ with respect to $x$ is $- \frac{6}{x^{4}}$ .

$- \frac{6}{x^{4}}$

Find the 4th Derivative f(x) = natural log of x

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