Calculus Examples

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

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 5}$ as ${(x + 5)}^{\frac{1}{2}}$ .

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = \frac{{(x + 5)}^{\frac{1}{2}}}{{(2 x^{2} - 4 x + 1)}^{4}}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $\frac{{(x + 5)}^{\frac{1}{2}}}{{(2 x^{2} - 4 x + 1)}^{4}}$ .

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

Step 2.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

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

Step 2.3

Replace all occurrences of $u_{1}$ with $\frac{{(x + 5)}^{\frac{1}{2}}}{{(2 x^{2} - 4 x + 1)}^{4}}$ .

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

Step 3

Multiply by the reciprocal of the fraction to divide by $\frac{{(x + 5)}^{\frac{1}{2}}}{{(2 x^{2} - 4 x + 1)}^{4}}$ .

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

Step 4

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

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

Step 5

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

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

Step 6

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

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Step 6.1

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

Step 6.2

Multiply $4$ by $2$ .

Step 7

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 $g (x) = x + 5$ .

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Step 7.1

To apply the Chain Rule, set $u_{2}$ as $x + 5$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{d}{d u_{2}} [{u_{2}}^{\frac{1}{2}}] \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 7.2

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {u_{2}}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 7.3

Replace all occurrences of $u_{2}$ with $x + 5$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {(x + 5)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 8

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {(x + 5)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 9

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {(x + 5)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 10

Combine the numerators over the common denominator.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {(x + 5)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 11

Simplify the numerator.

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Step 11.1

Multiply $- 1$ by $2$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {(x + 5)}^{\frac{1 - 2}{2}} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 11.2

Subtract $2$ from $1$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {(x + 5)}^{\frac{- 1}{2}} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 12

Combine fractions.

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Step 12.1

Move the negative in front of the fraction.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2} {(x + 5)}^{- \frac{1}{2}} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 12.2

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{{(x + 5)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 12.3

Move ${(x + 5)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} (\frac{1}{2 {(x + 5)}^{\frac{1}{2}}} \frac{d}{d x} [x + 5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 12.4

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} \frac{d}{d x} [x + 5] - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 13

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} (\frac{d}{d x} [x] + \frac{d}{d x} [5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 14

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} (1 + \frac{d}{d x} [5]) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 15

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} (1 + 0) - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 16

Simplify the expression.

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Step 16.1

Add $1$ and $0$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} \cdot 1 - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 16.2

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$

Step 17

Multiply $\frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}}$ by $\frac{2 {(x + 5)}^{\frac{1}{2}}}{2 {(x + 5)}^{\frac{1}{2}}}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} (\frac{2 {(x + 5)}^{\frac{1}{2}}}{2 {(x + 5)}^{\frac{1}{2}}} \cdot \frac{\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{{(2 x^{2} - 4 x + 1)}^{8}})$

Step 18

Combine.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{2 {(x + 5)}^{\frac{1}{2}} (\frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} - {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 19

Apply the distributive property.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{2 {(x + 5)}^{\frac{1}{2}} \frac{{(2 x^{2} - 4 x + 1)}^{4}}{2 {(x + 5)}^{\frac{1}{2}}} + 2 {(x + 5)}^{\frac{1}{2}} (- {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 20

Cancel the common factor of $2 {(x + 5)}^{\frac{1}{2}}$ .

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Step 20.1

Cancel the common factor.

Step 20.2

Rewrite the expression.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} + 2 {(x + 5)}^{\frac{1}{2}} (- {(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 21

Multiply $- 1$ by $2$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 {(x + 5)}^{\frac{1}{2}} ({(x + 5)}^{\frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 22

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 {(x + 5)}^{\frac{1}{2} + \frac{1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 23

Simplify the expression.

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Step 23.1

Combine the numerators over the common denominator.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 {(x + 5)}^{\frac{1 + 1}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 23.2

Add $1$ and $1$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 {(x + 5)}^{\frac{2}{2}} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 24

Cancel the common factor of $2$ .

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Step 24.1

Cancel the common factor.

Step 24.2

Rewrite the expression.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 {(x + 5)}^{1} \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 25

Simplify.

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 (x + 5) \frac{d}{d x} [{(2 x^{2} - 4 x + 1)}^{4}]}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 26

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 $g (x) = 2 x^{2} - 4 x + 1$ .

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Step 26.1

To apply the Chain Rule, set $u_{3}$ as $2 x^{2} - 4 x + 1$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 (x + 5) (\frac{d}{d u_{3}} [{u_{3}}^{4}] \frac{d}{d x} [2 x^{2} - 4 x + 1])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 26.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 4$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 (x + 5) (4 {u_{3}}^{3} \frac{d}{d x} [2 x^{2} - 4 x + 1])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 26.3

Replace all occurrences of $u_{3}$ with $2 x^{2} - 4 x + 1$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 2 (x + 5) (4 {(2 x^{2} - 4 x + 1)}^{3} \frac{d}{d x} [2 x^{2} - 4 x + 1])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27

Differentiate.

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Step 27.1

Multiply $4$ by $- 2$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} \frac{d}{d x} [2 x^{2} - 4 x + 1])}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.2

By the Sum Rule, the derivative of $2 x^{2} - 4 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.3

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.4

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (2 (2 x) + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.5

Multiply $2$ by $2$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.6

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4 \frac{d}{d x} [x] + \frac{d}{d x} [1]))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.7

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4 \cdot 1 + \frac{d}{d x} [1]))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.8

Multiply $- 4$ by $1$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4 + \frac{d}{d x} [1]))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.9

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4 + 0))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.10

Combine fractions.

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Step 27.10.1

Add $4 x - 4$ and $0$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}} \cdot \frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$

Step 27.10.2

Multiply $\frac{{(2 x^{2} - 4 x + 1)}^{4}}{{(x + 5)}^{\frac{1}{2}}}$ by $\frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4))}{2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8}}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(x + 5)}^{\frac{1}{2}} (2 {(x + 5)}^{\frac{1}{2}} {(2 x^{2} - 4 x + 1)}^{8})}$

Step 28

Multiply ${(x + 5)}^{\frac{1}{2}}$ by ${(x + 5)}^{\frac{1}{2}}$ by adding the exponents.

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Step 28.1

Move ${(x + 5)}^{\frac{1}{2}}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(x + 5)}^{\frac{1}{2}} {(x + 5)}^{\frac{1}{2}} (2 {(2 x^{2} - 4 x + 1)}^{8})}$

Step 28.2

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(x + 5)}^{\frac{1}{2} + \frac{1}{2}} (2 {(2 x^{2} - 4 x + 1)}^{8})}$

Step 28.3

Combine the numerators over the common denominator.

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(x + 5)}^{\frac{1 + 1}{2}} (2 {(2 x^{2} - 4 x + 1)}^{8})}$

Step 28.4

Add $1$ and $1$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(x + 5)}^{\frac{2}{2}} (2 {(2 x^{2} - 4 x + 1)}^{8})}$

Step 28.5

Divide $2$ by $2$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(x + 5)}^{1} (2 {(2 x^{2} - 4 x + 1)}^{8})}$

Step 29

Simplify ${(x + 5)}^{1} (2 {(2 x^{2} - 4 x + 1)}^{8})$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{(x + 5) (2 {(2 x^{2} - 4 x + 1)}^{8})}$

Step 30

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

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{2 \cdot (x + 5) {(2 x^{2} - 4 x + 1)}^{8}}$

Step 31

Cancel the common factors.

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Step 31.1

Factor ${(2 x^{2} - 4 x + 1)}^{4}$ out of $2 (x + 5) {(2 x^{2} - 4 x + 1)}^{8}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(2 x^{2} - 4 x + 1)}^{4} (2 (x + 5) {(2 x^{2} - 4 x + 1)}^{4})}$

Step 31.2

Cancel the common factor.

$\frac{{(2 x^{2} - 4 x + 1)}^{4} ({(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)))}{{(2 x^{2} - 4 x + 1)}^{4} (2 (x + 5) {(2 x^{2} - 4 x + 1)}^{4})}$

Step 31.3

Rewrite the expression.

$\frac{{(2 x^{2} - 4 x + 1)}^{4} - 8 (x + 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4))}{2 (x + 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32

Simplify.

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Step 32.1

Apply the distributive property.

$\frac{{(2 x^{2} - 4 x + 1)}^{4} + (- 8 x - 8 \cdot 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4))}{2 (x + 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.2

Apply the distributive property.

$\frac{{(2 x^{2} - 4 x + 1)}^{4} + (- 8 x - 8 \cdot 5) ({(2 x^{2} - 4 x + 1)}^{3} (4 x - 4))}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3

Simplify the numerator.

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Step 32.3.1

Factor ${(2 x^{2} - 4 x + 1)}^{3}$ out of ${(2 x^{2} - 4 x + 1)}^{4} + (- 8 x - 8 \cdot 5) {(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)$ .

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Step 32.3.1.1

Factor ${(2 x^{2} - 4 x + 1)}^{3}$ out of ${(2 x^{2} - 4 x + 1)}^{4}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1) + (- 8 x - 8 \cdot 5) {(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.1.2

Factor ${(2 x^{2} - 4 x + 1)}^{3}$ out of $(- 8 x - 8 \cdot 5) {(2 x^{2} - 4 x + 1)}^{3} (4 x - 4)$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1) + {(2 x^{2} - 4 x + 1)}^{3} ((- 8 x - 8 \cdot 5) (4 x - 4))}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.1.3

Factor ${(2 x^{2} - 4 x + 1)}^{3}$ out of ${(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1) + {(2 x^{2} - 4 x + 1)}^{3} ((- 8 x - 8 \cdot 5) (4 x - 4))$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 + (- 8 x - 8 \cdot 5) (4 x - 4))}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.2

Multiply $- 8$ by $5$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 + (- 8 x - 40) (4 x - 4))}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.3

Expand $(- 8 x - 40) (4 x - 4)$ using the FOIL Method.

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Step 32.3.3.1

Apply the distributive property.

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 8 x (4 x - 4) - 40 (4 x - 4))}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.3.2

Apply the distributive property.

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 8 x (4 x) - 8 x \cdot - 4 - 40 (4 x - 4))}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.3.3

Apply the distributive property.

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 8 x (4 x) - 8 x \cdot - 4 - 40 (4 x) - 40 \cdot - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4

Simplify and combine like terms.

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Step 32.3.4.1

Simplify each term.

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Step 32.3.4.1.1

Rewrite using the commutative property of multiplication.

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 8 \cdot 4 x \cdot x - 8 x \cdot - 4 - 40 (4 x) - 40 \cdot - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 32.3.4.1.2.1

Move $x$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 8 \cdot 4 (x \cdot x) - 8 x \cdot - 4 - 40 (4 x) - 40 \cdot - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4.1.2.2

Multiply $x$ by $x$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 8 \cdot 4 x^{2} - 8 x \cdot - 4 - 40 (4 x) - 40 \cdot - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4.1.3

Multiply $- 8$ by $4$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 32 x^{2} - 8 x \cdot - 4 - 40 (4 x) - 40 \cdot - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4.1.4

Multiply $- 4$ by $- 8$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 32 x^{2} + 32 x - 40 (4 x) - 40 \cdot - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4.1.5

Multiply $4$ by $- 40$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 32 x^{2} + 32 x - 160 x - 40 \cdot - 4)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4.1.6

Multiply $- 40$ by $- 4$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 32 x^{2} + 32 x - 160 x + 160)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.4.2

Subtract $160 x$ from $32 x$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (2 x^{2} - 4 x + 1 - 32 x^{2} - 128 x + 160)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.5

Subtract $32 x^{2}$ from $2 x^{2}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (- 30 x^{2} - 4 x + 1 - 128 x + 160)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.6

Subtract $128 x$ from $- 4 x$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (- 30 x^{2} - 132 x + 1 + 160)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.3.7

Add $1$ and $160$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (- 30 x^{2} - 132 x + 161)}{(2 x + 2 \cdot 5) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.4

Combine terms.

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Step 32.4.1

Multiply $2$ by $5$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (- 30 x^{2} - 132 x + 161)}{(2 x + 10) {(2 x^{2} - 4 x + 1)}^{4}}$

Step 32.4.2

Cancel the common factors.

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Step 32.4.2.1

Factor ${(2 x^{2} - 4 x + 1)}^{3}$ out of $(2 x + 10) {(2 x^{2} - 4 x + 1)}^{4}$ .

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (- 30 x^{2} - 132 x + 161)}{{(2 x^{2} - 4 x + 1)}^{3} ((2 x + 10) (2 x^{2} - 4 x + 1))}$

Step 32.4.2.2

Cancel the common factor.

$\frac{{(2 x^{2} - 4 x + 1)}^{3} (- 30 x^{2} - 132 x + 161)}{{(2 x^{2} - 4 x + 1)}^{3} ((2 x + 10) (2 x^{2} - 4 x + 1))}$

Step 32.4.2.3

Rewrite the expression.

$\frac{- 30 x^{2} - 132 x + 161}{(2 x + 10) (2 x^{2} - 4 x + 1)}$

Step 32.5

Factor $2$ out of $2 x + 10$ .

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Step 32.5.1

Factor $2$ out of $2 x$ .

$\frac{- 30 x^{2} - 132 x + 161}{(2 (x) + 10) (2 x^{2} - 4 x + 1)}$

Step 32.5.2

Factor $2$ out of $10$ .

$\frac{- 30 x^{2} - 132 x + 161}{(2 x + 2 \cdot 5) (2 x^{2} - 4 x + 1)}$

Step 32.5.3

Factor $2$ out of $2 x + 2 \cdot 5$ .

$\frac{- 30 x^{2} - 132 x + 161}{2 (x + 5) (2 x^{2} - 4 x + 1)}$

Step 32.6

Factor $- 1$ out of $- 30 x^{2}$ .

$\frac{- (30 x^{2}) - 132 x + 161}{2 (x + 5) (2 x^{2} - 4 x + 1)}$

Step 32.7

Factor $- 1$ out of $- 132 x$ .

$\frac{- (30 x^{2}) - (132 x) + 161}{2 (x + 5) (2 x^{2} - 4 x + 1)}$

Step 32.8

Factor $- 1$ out of $- (30 x^{2}) - (132 x)$ .

$\frac{- (30 x^{2} + 132 x) + 161}{2 (x + 5) (2 x^{2} - 4 x + 1)}$

Step 32.9

Rewrite $161$ as $- 1 (- 161)$ .

$\frac{- (30 x^{2} + 132 x) - 1 (- 161)}{2 (x + 5) (2 x^{2} - 4 x + 1)}$

Step 32.10

Factor $- 1$ out of $- (30 x^{2} + 132 x) - 1 (- 161)$ .

$\frac{- (30 x^{2} + 132 x - 161)}{2 (x + 5) (2 x^{2} - 4 x + 1)}$

Step 32.11

Rewrite $- (30 x^{2} + 132 x - 161)$ as $- 1 (30 x^{2} + 132 x - 161)$ .

$\frac{- 1 (30 x^{2} + 132 x - 161)}{2 (x + 5) (2 x^{2} - 4 x + 1)}$

Step 32.12

Move the negative in front of the fraction.

$- \frac{30 x^{2} + 132 x - 161}{2 (x + 5) (2 x^{2} - 4 x + 1)}$