Calculus Examples

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

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{2}{3}}$ and $g (x) = x + 3$ .

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Step 1.1.1.1

To apply the Chain Rule, set $u$ as $x + 3$ .

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

Step 1.1.1.2

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

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

Step 1.1.1.3

Replace all occurrences of $u$ with $x + 3$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

Combine the numerators over the common denominator.

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

Step 1.1.5

Simplify the numerator.

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Step 1.1.5.1

Multiply $- 1$ by $3$ .

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

Step 1.1.5.2

Subtract $3$ from $2$ .

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

Step 1.1.6

Combine fractions.

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Step 1.1.6.1

Move the negative in front of the fraction.

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

Step 1.1.6.2

Combine $\frac{2}{3}$ and ${(x + 3)}^{- \frac{1}{3}}$ .

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

Step 1.1.6.3

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

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

Step 1.1.7

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

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

Step 1.1.8

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

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

Step 1.1.9

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

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

Step 1.1.10

Simplify the expression.

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Step 1.1.10.1

Add $1$ and $0$ .

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

Step 1.1.10.2

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

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

Step 1.2

Find the second derivative.

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Step 1.2.1

Differentiate using the Constant Multiple Rule.

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Step 1.2.1.1

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

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

Step 1.2.1.2

Apply basic rules of exponents.

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Step 1.2.1.2.1

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

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

Step 1.2.1.2.2

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

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Step 1.2.1.2.2.1

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

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

Step 1.2.1.2.2.2

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

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

Step 1.2.1.2.2.3

Move the negative in front of the fraction.

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

Step 1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{1}{3}}$ and $g (x) = x + 3$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $x + 3$ .

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

Step 1.2.2.2

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

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

Step 1.2.2.3

Replace all occurrences of $u$ with $x + 3$ .

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

Combine the numerators over the common denominator.

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

Step 1.2.6

Simplify the numerator.

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Step 1.2.6.1

Multiply $- 1$ by $3$ .

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

Step 1.2.6.2

Subtract $3$ from $- 1$ .

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

Step 1.2.7

Combine fractions.

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Step 1.2.7.1

Move the negative in front of the fraction.

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

Step 1.2.7.2

Combine ${(x + 3)}^{- \frac{4}{3}}$ and $\frac{1}{3}$ .

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

Step 1.2.7.3

Move ${(x + 3)}^{- \frac{4}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.2.7.4

Multiply $\frac{2}{3}$ by $\frac{1}{3 {(x + 3)}^{\frac{4}{3}}}$ .

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

Step 1.2.7.5

Multiply $3$ by $3$ .

$f'' (x) = - \frac{2}{9 {(x + 3)}^{\frac{4}{3}}} \cdot \frac{d}{d x} (x + 3)$

Step 1.2.8

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$f'' (x) = - \frac{2}{9 {(x + 3)}^{\frac{4}{3}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (3))$

Step 1.2.9

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

$f'' (x) = - \frac{2}{9 {(x + 3)}^{\frac{4}{3}}} \cdot (1 + \frac{d}{d x} (3))$

Step 1.2.10

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{2}{9 {(x + 3)}^{\frac{4}{3}}} \cdot (1 + 0)$

Step 1.2.11

Simplify the expression.

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Step 1.2.11.1

Add $1$ and $0$ .

$f'' (x) = - \frac{2}{9 {(x + 3)}^{\frac{4}{3}}} \cdot 1$

Step 1.2.11.2

Multiply $- 1$ by $1$ .

$f'' (x) = - \frac{2}{9 {(x + 3)}^{\frac{4}{3}}}$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{2}{9 {(x + 3)}^{\frac{4}{3}}}$ .

$- \frac{2}{9 {(x + 3)}^{\frac{4}{3}}}$

Step 2

Set the second derivative equal to $0$ then solve the equation $- \frac{2}{9 {(x + 3)}^{\frac{4}{3}}} = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$- \frac{2}{9 {(x + 3)}^{\frac{4}{3}}} = 0$

Step 2.2

Set the numerator equal to zero.

$2 = 0$

Step 2.3

Since $2 \neq 0$ , there are no solutions.

No solution

Step 3

No values found that can make the second derivative equal to $0$ .

No Inflection Points