Calculus Examples

$y = \log_{2} ({(2 x^{2} - x)}^{\frac{5}{2}})$

Step 1

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

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Step 1.1

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

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

Step 1.2

The derivative of $\log_{2} (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1} \ln (2)}$ .

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

Step 1.3

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = \frac{5}{2}$ .

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

Step 2.3

Replace all occurrences of $u_{2}$ with $2 x^{2} - x$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $5$ .

$\frac{1}{{(2 x^{2} - x)}^{\frac{5}{2}} \ln (2)} (\frac{5}{2} {(2 x^{2} - x)}^{\frac{3}{2}} \frac{d}{d x} [2 x^{2} - x])$

Step 7

Combine $\frac{5}{2}$ and ${(2 x^{2} - x)}^{\frac{3}{2}}$ .

$\frac{1}{{(2 x^{2} - x)}^{\frac{5}{2}} \ln (2)} (\frac{5 {(2 x^{2} - x)}^{\frac{3}{2}}}{2} \frac{d}{d x} [2 x^{2} - x])$

Step 8

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

$\frac{5 {(2 x^{2} - x)}^{\frac{3}{2}}}{2 ({(2 x^{2} - x)}^{\frac{5}{2}} \ln (2))} \frac{d}{d x} [2 x^{2} - x]$

Step 9

Move ${(2 x^{2} - x)}^{\frac{3}{2}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

$\frac{5}{2 {(2 x^{2} - x)}^{\frac{5}{2}} \ln (2) {(2 x^{2} - x)}^{- \frac{3}{2}}} \frac{d}{d x} [2 x^{2} - x]$

Step 10

Simplify the denominator.

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Step 10.1

Multiply ${(2 x^{2} - x)}^{\frac{5}{2}}$ by ${(2 x^{2} - x)}^{- \frac{3}{2}}$ by adding the exponents.

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Step 10.1.1

Move ${(2 x^{2} - x)}^{- \frac{3}{2}}$ .

$\frac{5}{2 ({(2 x^{2} - x)}^{- \frac{3}{2}} {(2 x^{2} - x)}^{\frac{5}{2}}) \ln (2)} \frac{d}{d x} [2 x^{2} - x]$

Step 10.1.2

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

$\frac{5}{2 {(2 x^{2} - x)}^{- \frac{3}{2} + \frac{5}{2}} \ln (2)} \frac{d}{d x} [2 x^{2} - x]$

Step 10.1.3

Combine the numerators over the common denominator.

$\frac{5}{2 {(2 x^{2} - x)}^{\frac{- 3 + 5}{2}} \ln (2)} \frac{d}{d x} [2 x^{2} - x]$

Step 10.1.4

Add $- 3$ and $5$ .

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

Step 10.1.5

Divide $2$ by $2$ .

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

Step 10.2

Simplify $2 {(2 x^{2} - x)}^{1} \ln (2)$ .

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

Step 11

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

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

Step 12

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

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

Step 13

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

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

Step 14

Multiply $2$ by $2$ .

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

Step 15

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

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

Step 16

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

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

Step 17

Multiply $- 1$ by $1$ .

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

Step 18

Simplify.

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Step 18.1

Apply the distributive property.

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

Step 18.2

Apply the distributive property.

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

Step 18.3

Combine terms.

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Step 18.3.1

Multiply $2$ by $2$ .

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

Step 18.3.2

Multiply $- 1$ by $2$ .

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

Step 18.4

Reorder the factors of $\frac{5}{4 x^{2} \ln (2) - 2 x \ln (2)} (4 x - 1)$ .

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

Step 18.5

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

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Step 18.5.1

Factor $2 x \ln (2)$ out of $4 x^{2} \ln (2)$ .

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

Step 18.5.2

Factor $2 x \ln (2)$ out of $- 2 x \ln (2)$ .

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

Step 18.5.3

Factor $2 x \ln (2)$ out of $2 x \ln (2) (2 x) + 2 x \ln (2) (- 1)$ .

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

Step 18.6

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

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

Step 18.7

Move $5$ to the left of $4 x - 1$ .

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