Calculus Examples

$\sqrt[3]{x}$

Find the first derivative.

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ .

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

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

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Combine $- 1$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $- 1$ by $3$ .

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

Subtract $3$ from $1$ .

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

Move the negative in front of the fraction.

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

Simplify.

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

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

Multiply $\frac{1}{3}$ by $\frac{1}{x^{\frac{2}{3}}}$ .

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

Find the second derivative.

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

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

Apply basic rules of exponents.

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Rewrite $\frac{1}{x^{\frac{2}{3}}}$ as ${(x^{\frac{2}{3}})}^{- 1}$ .

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

Multiply the exponents in ${(x^{\frac{2}{3}})}^{- 1}$ .

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

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

Multiply $\frac{2}{3} \cdot - 1$ .

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Combine $\frac{2}{3}$ and $- 1$ .

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - \frac{2}{3}$ .

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Combine $- 1$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $- 1$ by $3$ .

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

Subtract $3$ from $- 2$ .

$f'' (x) = \frac{1}{3} \cdot (- \frac{2}{3} x^{\frac{- 5}{3}})$

Move the negative in front of the fraction.

$f'' (x) = \frac{1}{3} \cdot (- \frac{2}{3} x^{- \frac{5}{3}})$

Combine $x^{- \frac{5}{3}}$ and $\frac{2}{3}$ .

$f'' (x) = \frac{1}{3} \cdot (- \frac{x^{- \frac{5}{3}} \cdot 2}{3})$

Multiply $\frac{1}{3}$ by $\frac{x^{- \frac{5}{3}} \cdot 2}{3}$ .

$f'' (x) = - \frac{x^{- \frac{5}{3}} \cdot 2}{3 \cdot 3}$

Simplify the expression.

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Multiply $3$ by $3$ .

$f'' (x) = - \frac{x^{- \frac{5}{3}} \cdot 2}{9}$

Move $2$ to the left of $x^{- \frac{5}{3}}$ .

$f'' (x) = - \frac{2 \cdot x^{- \frac{5}{3}}}{9}$

Move $x^{- \frac{5}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = - \frac{2}{9 x^{\frac{5}{3}}}$