Calculus Examples

$y = \frac{3 x - 2}{2 x + 1}$

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 $f (x) = 3 x - 2$ and $g (x) = 2 x + 1$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $3 x - 2$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 2]$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Multiply $3$ by $1$ .

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

Step 1.2.5

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Add $3$ and $0$ .

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

Step 1.2.6.2

Move $3$ to the left of $2 x + 1$ .

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

Step 1.2.7

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 1.2.8

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

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

Step 1.2.9

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

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

Step 1.2.10

Multiply $2$ by $1$ .

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

Step 1.2.11

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

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

Step 1.2.12

Simplify the expression.

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Step 1.2.12.1

Add $2$ and $0$ .

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

Step 1.2.12.2

Multiply $2$ by $- 1$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

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

Step 1.3.2

Apply the distributive property.

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

Step 1.3.3

Simplify the numerator.

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Step 1.3.3.1

Combine the opposite terms in $3 (2 x) + 3 \cdot 1 - 2 (3 x) - 2 \cdot - 2$ .

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Step 1.3.3.1.1

Reorder the factors in the terms $3 (2 x)$ and $- 2 (3 x)$ .

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

Step 1.3.3.1.2

Subtract $2 \cdot 3 x$ from $2 \cdot 3 x$ .

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

Step 1.3.3.1.3

Add $3 \cdot 1$ and $0$ .

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

Step 1.3.3.2

Simplify each term.

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Step 1.3.3.2.1

Multiply $3$ by $1$ .

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

Step 1.3.3.2.2

Multiply $- 2$ by $- 2$ .

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

Step 1.3.3.3

Add $3$ and $4$ .

$f' (x) = \frac{7}{{(2 x + 1)}^{2}}$

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 $7$ is constant with respect to $x$ , the derivative of $\frac{7}{{(2 x + 1)}^{2}}$ with respect to $x$ is $7 \frac{d}{d x} [\frac{1}{{(2 x + 1)}^{2}}]$ .

$f'' (x) = 7 \frac{d}{d x} (\frac{1}{{(2 x + 1)}^{2}})$

Step 2.1.2

Apply basic rules of exponents.

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Step 2.1.2.1

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

$f'' (x) = 7 \frac{d}{d x} ({({(2 x + 1)}^{2})}^{- 1})$

Step 2.1.2.2

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

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Step 2.1.2.2.1

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

$f'' (x) = 7 \frac{d}{d x} ({(2 x + 1)}^{2 \cdot - 1})$

Step 2.1.2.2.2

Multiply $2$ by $- 1$ .

$f'' (x) = 7 \frac{d}{d x} ({(2 x + 1)}^{- 2})$

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 $g (x) = 2 x + 1$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

$f'' (x) = 7 (\frac{d}{d u} (u^{- 2}) \frac{d}{d x} (2 x + 1))$

Step 2.2.2

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

$f'' (x) = 7 (- 2 u^{- 3} \frac{d}{d x} (2 x + 1))$

Step 2.2.3

Replace all occurrences of $u$ with $2 x + 1$ .

$f'' (x) = 7 (- 2 {(2 x + 1)}^{- 3} \frac{d}{d x} (2 x + 1))$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 2$ by $7$ .

$f'' (x) = - 14 ({(2 x + 1)}^{- 3} \frac{d}{d x} (2 x + 1))$

Step 2.3.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$f'' (x) = - 14 {(2 x + 1)}^{- 3} (\frac{d}{d x} (2 x) + \frac{d}{d x} (1))$

Step 2.3.3

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

$f'' (x) = - 14 {(2 x + 1)}^{- 3} (2 \frac{d}{d x} (x) + \frac{d}{d x} (1))$

Step 2.3.4

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

$f'' (x) = - 14 {(2 x + 1)}^{- 3} (2 \cdot 1 + \frac{d}{d x} (1))$

Step 2.3.5

Multiply $2$ by $1$ .

$f'' (x) = - 14 {(2 x + 1)}^{- 3} (2 + \frac{d}{d x} (1))$

Step 2.3.6

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

$f'' (x) = - 14 {(2 x + 1)}^{- 3} (2 + 0)$

Step 2.3.7

Simplify the expression.

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Step 2.3.7.1

Add $2$ and $0$ .

$f'' (x) = - 14 {(2 x + 1)}^{- 3} \cdot 2$

Step 2.3.7.2

Multiply $2$ by $- 14$ .

$f'' (x) = - 28 {(2 x + 1)}^{- 3}$

Step 2.4

Simplify.

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Step 2.4.1

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

$f'' (x) = - 28 \frac{1}{{(2 x + 1)}^{3}}$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $- 28$ and $\frac{1}{{(2 x + 1)}^{3}}$ .

$f'' (x) = \frac{- 28}{{(2 x + 1)}^{3}}$

Step 2.4.2.2

Move the negative in front of the fraction.

$f'' (x) = - \frac{28}{{(2 x + 1)}^{3}}$

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 $- 28$ is constant with respect to $x$ , the derivative of $- \frac{28}{{(2 x + 1)}^{3}}$ with respect to $x$ is $- 28 \frac{d}{d x} [\frac{1}{{(2 x + 1)}^{3}}]$ .

$f''' (x) = - 28 \frac{d}{d x} \frac{1}{{(2 x + 1)}^{3}}$

Step 3.1.2

Apply basic rules of exponents.

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Step 3.1.2.1

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

$f''' (x) = - 28 \frac{d}{d x} {({(2 x + 1)}^{3})}^{- 1}$

Step 3.1.2.2

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

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Step 3.1.2.2.1

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

$f''' (x) = - 28 \frac{d}{d x} {(2 x + 1)}^{3 \cdot - 1}$

Step 3.1.2.2.2

Multiply $3$ by $- 1$ .

$f''' (x) = - 28 \frac{d}{d x} {(2 x + 1)}^{- 3}$

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 $g (x) = 2 x + 1$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

$f''' (x) = - 28 (\frac{d}{d u} (u^{- 3}) \frac{d}{d x} (2 x + 1))$

Step 3.2.2

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

$f''' (x) = - 28 (- 3 u^{- 4} \frac{d}{d x} (2 x + 1))$

Step 3.2.3

Replace all occurrences of $u$ with $2 x + 1$ .

$f''' (x) = - 28 (- 3 {(2 x + 1)}^{- 4} \frac{d}{d x} (2 x + 1))$

Step 3.3

Differentiate.

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Step 3.3.1

Multiply $- 3$ by $- 28$ .

$f''' (x) = 84 ({(2 x + 1)}^{- 4} \frac{d}{d x} (2 x + 1))$

Step 3.3.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$f''' (x) = 84 {(2 x + 1)}^{- 4} (\frac{d}{d x} (2 x) + \frac{d}{d x} (1))$

Step 3.3.3

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

$f''' (x) = 84 {(2 x + 1)}^{- 4} (2 \frac{d}{d x} (x) + \frac{d}{d x} (1))$

Step 3.3.4

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

$f''' (x) = 84 {(2 x + 1)}^{- 4} (2 \cdot 1 + \frac{d}{d x} (1))$

Step 3.3.5

Multiply $2$ by $1$ .

$f''' (x) = 84 {(2 x + 1)}^{- 4} (2 + \frac{d}{d x} (1))$

Step 3.3.6

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

$f''' (x) = 84 {(2 x + 1)}^{- 4} (2 + 0)$

Step 3.3.7

Simplify the expression.

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Step 3.3.7.1

Add $2$ and $0$ .

$f''' (x) = 84 {(2 x + 1)}^{- 4} \cdot 2$

Step 3.3.7.2

Multiply $2$ by $84$ .

$f''' (x) = 168 {(2 x + 1)}^{- 4}$

Step 3.4

Simplify.

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Step 3.4.1

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

$f''' (x) = 168 (\frac{1}{{(2 x + 1)}^{4}})$

Step 3.4.2

Combine $168$ and $\frac{1}{{(2 x + 1)}^{4}}$ .

$f''' (x) = \frac{168}{{(2 x + 1)}^{4}}$

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 $168$ is constant with respect to $x$ , the derivative of $\frac{168}{{(2 x + 1)}^{4}}$ with respect to $x$ is $168 \frac{d}{d x} [\frac{1}{{(2 x + 1)}^{4}}]$ .

$f^{4} (x) = 168 \frac{d}{d x} (\frac{1}{{(2 x + 1)}^{4}})$

Step 4.1.2

Apply basic rules of exponents.

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Step 4.1.2.1

Rewrite $\frac{1}{{(2 x + 1)}^{4}}$ as ${({(2 x + 1)}^{4})}^{- 1}$ .

$f^{4} (x) = 168 \frac{d}{d x} ({({(2 x + 1)}^{4})}^{- 1})$

Step 4.1.2.2

Multiply the exponents in ${({(2 x + 1)}^{4})}^{- 1}$ .

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Step 4.1.2.2.1

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

$f^{4} (x) = 168 \frac{d}{d x} ({(2 x + 1)}^{4 \cdot - 1})$

Step 4.1.2.2.2

Multiply $4$ by $- 1$ .

$f^{4} (x) = 168 \frac{d}{d x} ({(2 x + 1)}^{- 4})$

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 $g (x) = 2 x + 1$ .

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Step 4.2.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

$f^{4} (x) = 168 (\frac{d}{d u} (u^{- 4}) \frac{d}{d x} (2 x + 1))$

Step 4.2.2

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

$f^{4} (x) = 168 (- 4 u^{- 5} \frac{d}{d x} (2 x + 1))$

Step 4.2.3

Replace all occurrences of $u$ with $2 x + 1$ .

$f^{4} (x) = 168 (- 4 {(2 x + 1)}^{- 5} \frac{d}{d x} (2 x + 1))$

Step 4.3

Differentiate.

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Step 4.3.1

Multiply $- 4$ by $168$ .

$f^{4} (x) = - 672 ({(2 x + 1)}^{- 5} \frac{d}{d x} (2 x + 1))$

Step 4.3.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$f^{4} (x) = - 672 {(2 x + 1)}^{- 5} (\frac{d}{d x} (2 x) + \frac{d}{d x} (1))$

Step 4.3.3

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

$f^{4} (x) = - 672 {(2 x + 1)}^{- 5} (2 \frac{d}{d x} (x) + \frac{d}{d x} (1))$

Step 4.3.4

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

$f^{4} (x) = - 672 {(2 x + 1)}^{- 5} (2 \cdot 1 + \frac{d}{d x} (1))$

Step 4.3.5

Multiply $2$ by $1$ .

$f^{4} (x) = - 672 {(2 x + 1)}^{- 5} (2 + \frac{d}{d x} (1))$

Step 4.3.6

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

$f^{4} (x) = - 672 {(2 x + 1)}^{- 5} (2 + 0)$

Step 4.3.7

Simplify the expression.

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Step 4.3.7.1

Add $2$ and $0$ .

$f^{4} (x) = - 672 {(2 x + 1)}^{- 5} \cdot 2$

Step 4.3.7.2

Multiply $2$ by $- 672$ .

$f^{4} (x) = - 1344 {(2 x + 1)}^{- 5}$

Step 4.4

Simplify.

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Step 4.4.1

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

$f^{4} (x) = - 1344 \frac{1}{{(2 x + 1)}^{5}}$

Step 4.4.2

Combine terms.

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Step 4.4.2.1

Combine $- 1344$ and $\frac{1}{{(2 x + 1)}^{5}}$ .

$f^{4} (x) = \frac{- 1344}{{(2 x + 1)}^{5}}$

Step 4.4.2.2

Move the negative in front of the fraction.

$f^{4} (x) = - \frac{1344}{{(2 x + 1)}^{5}}$