Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x + 1$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Multiply $x + 1$ by $1$ .

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

Simplify by adding terms.

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Step 1.2.6.1

Add $1$ and $0$ .

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

Step 1.2.6.2

Multiply $- 1$ by $1$ .

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

Step 1.2.6.3

Subtract $x$ from $x$ .

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

Step 1.2.6.4

Add $0$ and $1$ .

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

Step 2

Find the second derivative.

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Step 2.1

Apply basic rules of exponents.

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Step 2.1.1

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

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

Step 2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 2}$ and $g (x) = x + 1$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Simplify the expression.

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Step 2.3.4.1

Add $1$ and $0$ .

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

Step 2.3.4.2

Multiply $- 2$ by $1$ .

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

Step 2.4

Simplify.

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Step 2.4.1

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

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

Step 2.4.2

Combine terms.

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Step 2.4.2.1

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

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

Step 2.4.2.2

Move the negative in front of the fraction.

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

Step 3

Find the third derivative.

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Step 3.1

Differentiate using the Constant Multiple Rule.

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Step 3.1.1

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

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

Step 3.1.2

Apply basic rules of exponents.

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Step 3.1.2.1

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

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

Step 3.1.2.2

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

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Step 3.1.2.2.1

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

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

Step 3.1.2.2.2

Multiply $3$ by $- 1$ .

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

Step 3.2

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

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Step 3.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 3.2.2

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

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

Step 3.2.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 3.3

Differentiate.

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Step 3.3.1

Multiply $- 3$ by $- 2$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

$f''' (x) = 6 {(x + 1)}^{- 4} (1 + 0)$

Step 3.3.5

Simplify the expression.

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Step 3.3.5.1

Add $1$ and $0$ .

$f''' (x) = 6 {(x + 1)}^{- 4} \cdot 1$

Step 3.3.5.2

Multiply $6$ by $1$ .

$f''' (x) = 6 {(x + 1)}^{- 4}$

Step 3.4

Simplify.

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Step 3.4.1

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

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

Step 3.4.2

Combine $6$ and $\frac{1}{{(x + 1)}^{4}}$ .

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

Step 4

Find the fourth derivative.

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Step 4.1

Differentiate using the Constant Multiple Rule.

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Step 4.1.1

Since $6$ is constant with respect to $x$ , the derivative of $\frac{6}{{(x + 1)}^{4}}$ with respect to $x$ is $6 \frac{d}{d x} [\frac{1}{{(x + 1)}^{4}}]$ .

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

Step 4.1.2

Apply basic rules of exponents.

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Step 4.1.2.1

Rewrite $\frac{1}{{(x + 1)}^{4}}$ as ${({(x + 1)}^{4})}^{- 1}$ .

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

Step 4.1.2.2

Multiply the exponents in ${({(x + 1)}^{4})}^{- 1}$ .

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Step 4.1.2.2.1

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

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

Step 4.1.2.2.2

Multiply $4$ by $- 1$ .

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

Step 4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 4}$ and $g (x) = x + 1$ .

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Step 4.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 4$ .

$f^{4} (x) = 6 (- 4 u^{- 5} \frac{d}{d x} (x + 1))$

Step 4.2.3

Replace all occurrences of $u$ with $x + 1$ .

$f^{4} (x) = 6 (- 4 {(x + 1)}^{- 5} \frac{d}{d x} (x + 1))$

Step 4.3

Differentiate.

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Step 4.3.1

Multiply $- 4$ by $6$ .

$f^{4} (x) = - 24 ({(x + 1)}^{- 5} \frac{d}{d x} (x + 1))$

Step 4.3.2

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

$f^{4} (x) = - 24 {(x + 1)}^{- 5} (\frac{d}{d x} (x) + \frac{d}{d x} (1))$

Step 4.3.3

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

$f^{4} (x) = - 24 {(x + 1)}^{- 5} (1 + \frac{d}{d x} (1))$

Step 4.3.4

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

$f^{4} (x) = - 24 {(x + 1)}^{- 5} (1 + 0)$

Step 4.3.5

Simplify the expression.

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Step 4.3.5.1

Add $1$ and $0$ .

$f^{4} (x) = - 24 {(x + 1)}^{- 5} \cdot 1$

Step 4.3.5.2

Multiply $- 24$ by $1$ .

$f^{4} (x) = - 24 {(x + 1)}^{- 5}$

Step 4.4

Simplify.

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Step 4.4.1

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

$f^{4} (x) = - 24 \frac{1}{{(x + 1)}^{5}}$

Step 4.4.2

Combine terms.

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Step 4.4.2.1

Combine $- 24$ and $\frac{1}{{(x + 1)}^{5}}$ .

$f^{4} (x) = \frac{- 24}{{(x + 1)}^{5}}$

Step 4.4.2.2

Move the negative in front of the fraction.

$f^{4} (x) = - \frac{24}{{(x + 1)}^{5}}$

Step 5

The fourth derivative of $f (x)$ with respect to $x$ is $- \frac{24}{{(x + 1)}^{5}}$ .

$- \frac{24}{{(x + 1)}^{5}}$