Calculus Examples

$y = (2 x + 5) \sqrt{4 x - 1}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 x - 1}$ as ${(4 x - 1)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [(2 x + 5) {(4 x - 1)}^{\frac{1}{2}}]$

Step 2

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

$(2 x + 5) \frac{d}{d x} [{(4 x - 1)}^{\frac{1}{2}}] + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = 4 x - 1$ .

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Step 3.1

To apply the Chain Rule, set $u$ as $4 x - 1$ .

$(2 x + 5) (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 3.2

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

$(2 x + 5) (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 3.3

Replace all occurrences of $u$ with $4 x - 1$ .

$(2 x + 5) (\frac{1}{2} {(4 x - 1)}^{\frac{1}{2} - 1} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$(2 x + 5) (\frac{1}{2} {(4 x - 1)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 5

Combine $- 1$ and $\frac{2}{2}$ .

$(2 x + 5) (\frac{1}{2} {(4 x - 1)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 6

Combine the numerators over the common denominator.

$(2 x + 5) (\frac{1}{2} {(4 x - 1)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

$(2 x + 5) (\frac{1}{2} {(4 x - 1)}^{\frac{1 - 2}{2}} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 7.2

Subtract $2$ from $1$ .

$(2 x + 5) (\frac{1}{2} {(4 x - 1)}^{\frac{- 1}{2}} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 8

Combine fractions.

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Step 8.1

Move the negative in front of the fraction.

$(2 x + 5) (\frac{1}{2} {(4 x - 1)}^{- \frac{1}{2}} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 8.2

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

$(2 x + 5) (\frac{{(4 x - 1)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 8.3

Move ${(4 x - 1)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$(2 x + 5) (\frac{1}{2 {(4 x - 1)}^{\frac{1}{2}}} \frac{d}{d x} [4 x - 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 9

By the Sum Rule, the derivative of $4 x - 1$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 1]$ .

$(2 x + 5) \frac{1}{2 {(4 x - 1)}^{\frac{1}{2}}} (\frac{d}{d x} [4 x] + \frac{d}{d x} [- 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 10

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$(2 x + 5) \frac{1}{2 {(4 x - 1)}^{\frac{1}{2}}} (4 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 11

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

$(2 x + 5) \frac{1}{2 {(4 x - 1)}^{\frac{1}{2}}} (4 \cdot 1 + \frac{d}{d x} [- 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 12

Multiply $4$ by $1$ .

$(2 x + 5) \frac{1}{2 {(4 x - 1)}^{\frac{1}{2}}} (4 + \frac{d}{d x} [- 1]) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 13

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

$(2 x + 5) \frac{1}{2 {(4 x - 1)}^{\frac{1}{2}}} (4 + 0) + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 14

Simplify terms.

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Step 14.1

Add $4$ and $0$ .

$(2 x + 5) \frac{1}{2 {(4 x - 1)}^{\frac{1}{2}}} \cdot 4 + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 14.2

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

$(2 x + 5) \frac{4}{2 {(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 14.3

Factor $2$ out of $4$ .

$(2 x + 5) \frac{2 \cdot 2}{2 {(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 15

Cancel the common factors.

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Step 15.1

Factor $2$ out of $2 {(4 x - 1)}^{\frac{1}{2}}$ .

$(2 x + 5) \frac{2 \cdot 2}{2 ({(4 x - 1)}^{\frac{1}{2}})} + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 15.2

Cancel the common factor.

$(2 x + 5) \frac{2 \cdot 2}{2 {(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 15.3

Rewrite the expression.

$(2 x + 5) \frac{2}{{(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} \frac{d}{d x} [2 x + 5]$

Step 16

By the Sum Rule, the derivative of $2 x + 5$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [5]$ .

$(2 x + 5) \frac{2}{{(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} (\frac{d}{d x} [2 x] + \frac{d}{d x} [5])$

Step 17

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

$(2 x + 5) \frac{2}{{(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} (2 \frac{d}{d x} [x] + \frac{d}{d x} [5])$

Step 18

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

$(2 x + 5) \frac{2}{{(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} (2 \cdot 1 + \frac{d}{d x} [5])$

Step 19

Multiply $2$ by $1$ .

$(2 x + 5) \frac{2}{{(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} (2 + \frac{d}{d x} [5])$

Step 20

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$(2 x + 5) \frac{2}{{(4 x - 1)}^{\frac{1}{2}}} + {(4 x - 1)}^{\frac{1}{2}} (2 + 0)$

Step 21

Simplify the expression.

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Step 21.1

Add $2$ and $0$ .

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

Step 21.2

Move $2$ to the left of ${(4 x - 1)}^{\frac{1}{2}}$ .

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

Step 22

Simplify.

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Step 22.1

Simplify each term.

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Step 22.1.1

Multiply $2 x + 5$ by $\frac{2}{{(4 x - 1)}^{\frac{1}{2}}}$ .

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

Step 22.1.2

Move $2$ to the left of $2 x + 5$ .

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

Step 22.2

To write $2 {(4 x - 1)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{{(4 x - 1)}^{\frac{1}{2}}}{{(4 x - 1)}^{\frac{1}{2}}}$ .

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

Step 22.3

Combine the numerators over the common denominator.

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

Step 22.4

Simplify the numerator.

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Step 22.4.1

Factor $2$ out of $2 (2 x + 5) + 2 {(4 x - 1)}^{\frac{1}{2}} {(4 x - 1)}^{\frac{1}{2}}$ .

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Step 22.4.1.1

Factor $2$ out of $2 {(4 x - 1)}^{\frac{1}{2}} {(4 x - 1)}^{\frac{1}{2}}$ .

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

Step 22.4.1.2

Factor $2$ out of $2 (2 x + 5) + 2 ({(4 x - 1)}^{\frac{1}{2}} {(4 x - 1)}^{\frac{1}{2}})$ .

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

Step 22.4.2

Multiply ${(4 x - 1)}^{\frac{1}{2}}$ by ${(4 x - 1)}^{\frac{1}{2}}$ by adding the exponents.

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Step 22.4.2.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 22.4.2.2

Combine the numerators over the common denominator.

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

Step 22.4.2.3

Add $1$ and $1$ .

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

Step 22.4.2.4

Divide $2$ by $2$ .

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

Step 22.4.3

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

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

Step 22.4.4

Add $2 x$ and $4 x$ .

$\frac{2 (6 x + 5 - 1)}{{(4 x - 1)}^{\frac{1}{2}}}$

Step 22.4.5

Subtract $1$ from $5$ .

$\frac{2 (6 x + 4)}{{(4 x - 1)}^{\frac{1}{2}}}$

Step 22.4.6

Factor $2$ out of $6 x + 4$ .

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Step 22.4.6.1

Factor $2$ out of $6 x$ .

$\frac{2 (2 (3 x) + 4)}{{(4 x - 1)}^{\frac{1}{2}}}$

Step 22.4.6.2

Factor $2$ out of $4$ .

$\frac{2 (2 (3 x) + 2 (2))}{{(4 x - 1)}^{\frac{1}{2}}}$

Step 22.4.6.3

Factor $2$ out of $2 (3 x) + 2 (2)$ .

$\frac{2 (2 (3 x + 2))}{{(4 x - 1)}^{\frac{1}{2}}}$

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

Step 22.4.7

Multiply $2$ by $2$ .

$\frac{4 (3 x + 2)}{{(4 x - 1)}^{\frac{1}{2}}}$