Calculus Examples

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

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

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

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

Step 1.1.2

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

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

Step 1.1.3

Differentiate.

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Step 1.1.3.1

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

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

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

Step 1.1.3.4

Simplify the expression.

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Step 1.1.3.4.1

Add $1$ and $0$ .

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

Step 1.1.3.4.2

Multiply $- 1$ by $1$ .

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

Step 1.1.4

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

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

Step 1.2

Find the second derivative.

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Step 1.2.1

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 $g (x) = \frac{1}{{(x + 3)}^{2}}$ .

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

Step 1.2.2

Apply basic rules of exponents.

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Step 1.2.2.1

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

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

Step 1.2.2.2

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

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Step 1.2.2.2.1

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

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

Step 1.2.2.2.2

Multiply $2$ by $- 1$ .

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

Step 1.2.3.2

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

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

Step 1.2.3.3

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

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

Step 1.2.4

Differentiate.

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Step 1.2.4.1

Multiply $- 2$ by $- 1$ .

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

Step 1.2.4.2

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

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

Step 1.2.4.3

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

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

Step 1.2.4.4

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

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

Step 1.2.4.5

Simplify the expression.

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Step 1.2.4.5.1

Add $1$ and $0$ .

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

Step 1.2.4.5.2

Multiply $2$ by $1$ .

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

Step 1.2.4.6

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

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

Step 1.2.4.7

Simplify the expression.

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Step 1.2.4.7.1

Multiply $\frac{1}{{(x + 3)}^{2}}$ by $0$ .

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

Step 1.2.4.7.2

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

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

Step 1.2.5

Simplify.

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Step 1.2.5.1

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

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

Step 1.2.5.2

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

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

Step 1.3

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

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

Step 2

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

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Step 2.1

Set the second derivative equal to $0$ .

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

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