Calculus Examples

$y = \frac{1 + x}{1 - x}$

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 $f (x) = 1 + x$ and $g (x) = 1 - x$ .

$\frac{(1 - x) \frac{d}{d x} [1 + x] - (1 + x) \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{(1 - x) (\frac{d}{d x} [1] + \frac{d}{d x} [x]) - (1 + x) \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.2.2

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

$\frac{(1 - x) (0 + \frac{d}{d x} [x]) - (1 + x) \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.2.3

Add $0$ and $\frac{d}{d x} [x]$ .

$\frac{(1 - x) \frac{d}{d x} [x] - (1 + x) \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.2.4

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

$\frac{(1 - x) \cdot 1 - (1 + x) \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.2.5

Multiply $1 - x$ by $1$ .

$\frac{1 - x - (1 + x) \frac{d}{d x} [1 - x]}{{(1 - x)}^{2}}$

Step 1.2.6

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

$\frac{1 - x - (1 + x) (\frac{d}{d x} [1] + \frac{d}{d x} [- x])}{{(1 - x)}^{2}}$

Step 1.2.7

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

$\frac{1 - x - (1 + x) (0 + \frac{d}{d x} [- x])}{{(1 - x)}^{2}}$

Step 1.2.8

Add $0$ and $\frac{d}{d x} [- x]$ .

$\frac{1 - x - (1 + x) \frac{d}{d x} [- x]}{{(1 - x)}^{2}}$

Step 1.2.9

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

$\frac{1 - x - (1 + x) (- \frac{d}{d x} [x])}{{(1 - x)}^{2}}$

Step 1.2.10

Multiply.

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Step 1.2.10.1

Multiply $- 1$ by $- 1$ .

$\frac{1 - x + 1 (1 + x) \frac{d}{d x} [x]}{{(1 - x)}^{2}}$

Step 1.2.10.2

Multiply $1 + x$ by $1$ .

$\frac{1 - x + (1 + x) \frac{d}{d x} [x]}{{(1 - x)}^{2}}$

Step 1.2.11

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

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

Step 1.2.12

Simplify by adding terms.

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Step 1.2.12.1

Multiply $1 + x$ by $1$ .

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

Step 1.2.12.2

Add $1$ and $1$ .

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

Step 1.2.12.3

Add $- x$ and $x$ .

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

Step 1.2.12.4

Simplify the expression.

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Step 1.2.12.4.1

Add $0$ and $2$ .

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

Step 1.2.12.4.2

Reorder terms.

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

Step 2

Find the second derivative.

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Step 2.1

Differentiate using the Constant Multiple Rule.

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Step 2.1.1

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

$2 \frac{d}{d x} [\frac{1}{{(- x + 1)}^{2}}]$

Step 2.1.2

Apply basic rules of exponents.

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Step 2.1.2.1

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

$2 \frac{d}{d x} [{({(- x + 1)}^{2})}^{- 1}]$

Step 2.1.2.2

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

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Step 2.1.2.2.1

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

$2 \frac{d}{d x} [{(- x + 1)}^{2 \cdot - 1}]$

Step 2.1.2.2.2

Multiply $2$ by $- 1$ .

$2 \frac{d}{d x} [{(- x + 1)}^{- 2}]$

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 $- x + 1$ .

$2 (\frac{d}{d u} [u^{- 2}] \frac{d}{d x} [- x + 1])$

Step 2.2.2

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

$2 (- 2 u^{- 3} \frac{d}{d x} [- x + 1])$

Step 2.2.3

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

$2 (- 2 {(- x + 1)}^{- 3} \frac{d}{d x} [- x + 1])$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 2$ by $2$ .

$- 4 ({(- x + 1)}^{- 3} \frac{d}{d x} [- x + 1])$

Step 2.3.2

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

$- 4 {(- x + 1)}^{- 3} (\frac{d}{d x} [- x] + \frac{d}{d x} [1])$

Step 2.3.3

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

$- 4 {(- x + 1)}^{- 3} (- \frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 2.3.4

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

$- 4 {(- x + 1)}^{- 3} (- 1 \cdot 1 + \frac{d}{d x} [1])$

Step 2.3.5

Multiply $- 1$ by $1$ .

$- 4 {(- x + 1)}^{- 3} (- 1 + \frac{d}{d x} [1])$

Step 2.3.6

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

$- 4 {(- x + 1)}^{- 3} (- 1 + 0)$

Step 2.3.7

Simplify the expression.

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Step 2.3.7.1

Add $- 1$ and $0$ .

$- 4 {(- x + 1)}^{- 3} \cdot - 1$

Step 2.3.7.2

Multiply $- 1$ by $- 4$ .

$4 {(- x + 1)}^{- 3}$

Step 2.4

Simplify.

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Step 2.4.1

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

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

Step 2.4.2

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

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

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 $4$ is constant with respect to $x$ , the derivative of $\frac{4}{{(- x + 1)}^{3}}$ with respect to $x$ is $4 \frac{d}{d x} [\frac{1}{{(- x + 1)}^{3}}]$ .

$4 \frac{d}{d x} [\frac{1}{{(- x + 1)}^{3}}]$

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 ${({(- x + 1)}^{3})}^{- 1}$ .

$4 \frac{d}{d x} [{({(- x + 1)}^{3})}^{- 1}]$

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, ${(a^{m})}^{n} = a^{m n}$ .

$4 \frac{d}{d x} [{(- x + 1)}^{3 \cdot - 1}]$

Step 3.1.2.2.2

Multiply $3$ by $- 1$ .

$4 \frac{d}{d x} [{(- x + 1)}^{- 3}]$

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 $- x + 1$ .

$4 (\frac{d}{d u} [u^{- 3}] \frac{d}{d x} [- x + 1])$

Step 3.2.2

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

$4 (- 3 u^{- 4} \frac{d}{d x} [- x + 1])$

Step 3.2.3

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

$4 (- 3 {(- x + 1)}^{- 4} \frac{d}{d x} [- x + 1])$

Step 3.3

Differentiate.

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Step 3.3.1

Multiply $- 3$ by $4$ .

$- 12 ({(- x + 1)}^{- 4} \frac{d}{d x} [- x + 1])$

Step 3.3.2

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

$- 12 {(- x + 1)}^{- 4} (\frac{d}{d x} [- x] + \frac{d}{d x} [1])$

Step 3.3.3

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

$- 12 {(- x + 1)}^{- 4} (- \frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 3.3.4

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

$- 12 {(- x + 1)}^{- 4} (- 1 \cdot 1 + \frac{d}{d x} [1])$

Step 3.3.5

Multiply $- 1$ by $1$ .

$- 12 {(- x + 1)}^{- 4} (- 1 + \frac{d}{d x} [1])$

Step 3.3.6

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

$- 12 {(- x + 1)}^{- 4} (- 1 + 0)$

Step 3.3.7

Simplify the expression.

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Step 3.3.7.1

Add $- 1$ and $0$ .

$- 12 {(- x + 1)}^{- 4} \cdot - 1$

Step 3.3.7.2

Multiply $- 1$ by $- 12$ .

$12 {(- x + 1)}^{- 4}$

Step 3.4

Simplify.

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Step 3.4.1

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

$12 \frac{1}{{(- x + 1)}^{4}}$

Step 3.4.2

Combine $12$ and $\frac{1}{{(- x + 1)}^{4}}$ .

$f''' (x) = \frac{12}{{(- x + 1)}^{4}}$

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 $12$ is constant with respect to $x$ , the derivative of $\frac{12}{{(- x + 1)}^{4}}$ with respect to $x$ is $12 \frac{d}{d x} [\frac{1}{{(- x + 1)}^{4}}]$ .

$12 \frac{d}{d x} [\frac{1}{{(- x + 1)}^{4}}]$

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 ${({(- x + 1)}^{4})}^{- 1}$ .

$12 \frac{d}{d x} [{({(- x + 1)}^{4})}^{- 1}]$

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, ${(a^{m})}^{n} = a^{m n}$ .

$12 \frac{d}{d x} [{(- x + 1)}^{4 \cdot - 1}]$

Step 4.1.2.2.2

Multiply $4$ by $- 1$ .

$12 \frac{d}{d x} [{(- x + 1)}^{- 4}]$

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 $- x + 1$ .

$12 (\frac{d}{d u} [u^{- 4}] \frac{d}{d x} [- x + 1])$

Step 4.2.2

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

$12 (- 4 u^{- 5} \frac{d}{d x} [- x + 1])$

Step 4.2.3

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

$12 (- 4 {(- x + 1)}^{- 5} \frac{d}{d x} [- x + 1])$

Step 4.3

Differentiate.

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Step 4.3.1

Multiply $- 4$ by $12$ .

$- 48 ({(- x + 1)}^{- 5} \frac{d}{d x} [- x + 1])$

Step 4.3.2

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

$- 48 {(- x + 1)}^{- 5} (\frac{d}{d x} [- x] + \frac{d}{d x} [1])$

Step 4.3.3

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

$- 48 {(- x + 1)}^{- 5} (- \frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 4.3.4

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

$- 48 {(- x + 1)}^{- 5} (- 1 \cdot 1 + \frac{d}{d x} [1])$

Step 4.3.5

Multiply $- 1$ by $1$ .

$- 48 {(- x + 1)}^{- 5} (- 1 + \frac{d}{d x} [1])$

Step 4.3.6

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

$- 48 {(- x + 1)}^{- 5} (- 1 + 0)$

Step 4.3.7

Simplify the expression.

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Step 4.3.7.1

Add $- 1$ and $0$ .

$- 48 {(- x + 1)}^{- 5} \cdot - 1$

Step 4.3.7.2

Multiply $- 1$ by $- 48$ .

$48 {(- x + 1)}^{- 5}$

Step 4.4

Simplify.

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Step 4.4.1

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

$48 \frac{1}{{(- x + 1)}^{5}}$

Step 4.4.2

Combine $48$ and $\frac{1}{{(- x + 1)}^{5}}$ .

$f^{4} (x) = \frac{48}{{(- x + 1)}^{5}}$