Calculus Examples

${(2 x - 1)}^{3} {(x + 7)}^{- 3}$

Step 1

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 $f (x) = {(2 x - 1)}^{3}$ and $g (x) = {(x + 7)}^{- 3}$ .

${(2 x - 1)}^{3} \frac{d}{d x} [{(x + 7)}^{- 3}] + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 3}$ and $g (x) = x + 7$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $x + 7$ .

${(2 x - 1)}^{3} (\frac{d}{d u_{1}} [{u_{1}}^{- 3}] \frac{d}{d x} [x + 7]) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 2.2

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

${(2 x - 1)}^{3} (- 3 {u_{1}}^{- 4} \frac{d}{d x} [x + 7]) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 2.3

Replace all occurrences of $u_{1}$ with $x + 7$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4} \frac{d}{d x} [x + 7]) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4} (\frac{d}{d x} [x] + \frac{d}{d x} [7])) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 3.2

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

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4} (1 + \frac{d}{d x} [7])) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 3.3

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4} (1 + 0)) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $1$ and $0$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4} \cdot 1) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 3.4.2

Multiply $- 3$ by $1$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + {(x + 7)}^{- 3} \frac{d}{d x} [{(2 x - 1)}^{3}]$

Step 4

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 4.1

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

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + {(x + 7)}^{- 3} (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [2 x - 1])$

Step 4.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + {(x + 7)}^{- 3} (3 {u_{2}}^{2} \frac{d}{d x} [2 x - 1])$

Step 4.3

Replace all occurrences of $u_{2}$ with $2 x - 1$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + {(x + 7)}^{- 3} (3 {(2 x - 1)}^{2} \frac{d}{d x} [2 x - 1])$

Step 5

Differentiate.

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Step 5.1

Move $3$ to the left of ${(x + 7)}^{- 3}$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 3 \cdot {(x + 7)}^{- 3} {(2 x - 1)}^{2} \frac{d}{d x} [2 x - 1]$

Step 5.2

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

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 3 {(x + 7)}^{- 3} {(2 x - 1)}^{2} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1])$

Step 5.3

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

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 3 {(x + 7)}^{- 3} {(2 x - 1)}^{2} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1])$

Step 5.4

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

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 3 {(x + 7)}^{- 3} {(2 x - 1)}^{2} (2 \cdot 1 + \frac{d}{d x} [- 1])$

Step 5.5

Multiply $2$ by $1$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 3 {(x + 7)}^{- 3} {(2 x - 1)}^{2} (2 + \frac{d}{d x} [- 1])$

Step 5.6

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

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 3 {(x + 7)}^{- 3} {(2 x - 1)}^{2} (2 + 0)$

Step 5.7

Simplify the expression.

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Step 5.7.1

Add $2$ and $0$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 3 {(x + 7)}^{- 3} {(2 x - 1)}^{2} \cdot 2$

Step 5.7.2

Multiply $2$ by $3$ .

${(2 x - 1)}^{3} (- 3 {(x + 7)}^{- 4}) + 6 {(x + 7)}^{- 3} {(2 x - 1)}^{2}$

Step 6

Simplify.

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Step 6.1

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

${(2 x - 1)}^{3} (- 3 \frac{1}{{(x + 7)}^{4}}) + 6 {(x + 7)}^{- 3} {(2 x - 1)}^{2}$

Step 6.2

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

${(2 x - 1)}^{3} (- 3 \frac{1}{{(x + 7)}^{4}}) + 6 \frac{1}{{(x + 7)}^{3}} {(2 x - 1)}^{2}$

Step 6.3

Combine terms.

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Step 6.3.1

Combine $- 3$ and $\frac{1}{{(x + 7)}^{4}}$ .

${(2 x - 1)}^{3} \frac{- 3}{{(x + 7)}^{4}} + 6 \frac{1}{{(x + 7)}^{3}} {(2 x - 1)}^{2}$

Step 6.3.2

Move the negative in front of the fraction.

${(2 x - 1)}^{3} (- \frac{3}{{(x + 7)}^{4}}) + 6 \frac{1}{{(x + 7)}^{3}} {(2 x - 1)}^{2}$

Step 6.3.3

Combine ${(2 x - 1)}^{3}$ and $\frac{3}{{(x + 7)}^{4}}$ .

$- \frac{{(2 x - 1)}^{3} \cdot 3}{{(x + 7)}^{4}} + 6 \frac{1}{{(x + 7)}^{3}} {(2 x - 1)}^{2}$

Step 6.3.4

Move $3$ to the left of ${(2 x - 1)}^{3}$ .

$- \frac{3 \cdot {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + 6 \frac{1}{{(x + 7)}^{3}} {(2 x - 1)}^{2}$

Step 6.3.5

Combine $6$ and $\frac{1}{{(x + 7)}^{3}}$ .

$- \frac{3 {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + \frac{6}{{(x + 7)}^{3}} {(2 x - 1)}^{2}$

Step 6.3.6

Combine $\frac{6}{{(x + 7)}^{3}}$ and ${(2 x - 1)}^{2}$ .

$- \frac{3 {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + \frac{6 {(2 x - 1)}^{2}}{{(x + 7)}^{3}}$

Step 6.3.7

To write $\frac{6 {(2 x - 1)}^{2}}{{(x + 7)}^{3}}$ as a fraction with a common denominator, multiply by $\frac{x + 7}{x + 7}$ .

$- \frac{3 {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + \frac{6 {(2 x - 1)}^{2}}{{(x + 7)}^{3}} \cdot \frac{x + 7}{x + 7}$

Step 6.3.8

Write each expression with a common denominator of ${(x + 7)}^{4}$ , by multiplying each by an appropriate factor of $1$ .

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Step 6.3.8.1

Multiply $\frac{6 {(2 x - 1)}^{2}}{{(x + 7)}^{3}}$ by $\frac{x + 7}{x + 7}$ .

$- \frac{3 {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + \frac{6 {(2 x - 1)}^{2} (x + 7)}{{(x + 7)}^{3} (x + 7)}$

Step 6.3.8.2

Multiply ${(x + 7)}^{3}$ by $x + 7$ by adding the exponents.

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Step 6.3.8.2.1

Multiply ${(x + 7)}^{3}$ by $x + 7$ .

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Step 6.3.8.2.1.1

Raise $x + 7$ to the power of $1$ .

$- \frac{3 {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + \frac{6 {(2 x - 1)}^{2} (x + 7)}{{(x + 7)}^{3} {(x + 7)}^{1}}$

Step 6.3.8.2.1.2

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

$- \frac{3 {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + \frac{6 {(2 x - 1)}^{2} (x + 7)}{{(x + 7)}^{3 + 1}}$

Step 6.3.8.2.2

Add $3$ and $1$ .

$- \frac{3 {(2 x - 1)}^{3}}{{(x + 7)}^{4}} + \frac{6 {(2 x - 1)}^{2} (x + 7)}{{(x + 7)}^{4}}$

Step 6.3.9

Combine the numerators over the common denominator.

$\frac{- 3 {(2 x - 1)}^{3} + 6 {(2 x - 1)}^{2} (x + 7)}{{(x + 7)}^{4}}$

Step 6.4

Simplify the numerator.

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Step 6.4.1

Factor $3 {(2 x - 1)}^{2}$ out of $- 3 {(2 x - 1)}^{3} + 6 {(2 x - 1)}^{2} (x + 7)$ .

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Step 6.4.1.1

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

$\frac{3 {(2 x - 1)}^{2} (- (2 x - 1)) + 6 {(2 x - 1)}^{2} (x + 7)}{{(x + 7)}^{4}}$

Step 6.4.1.2

Factor $3 {(2 x - 1)}^{2}$ out of $6 {(2 x - 1)}^{2} (x + 7)$ .

$\frac{3 {(2 x - 1)}^{2} (- (2 x - 1)) + 3 {(2 x - 1)}^{2} (2 (x + 7))}{{(x + 7)}^{4}}$

Step 6.4.1.3

Factor $3 {(2 x - 1)}^{2}$ out of $3 {(2 x - 1)}^{2} (- (2 x - 1)) + 3 {(2 x - 1)}^{2} (2 (x + 7))$ .

$\frac{3 {(2 x - 1)}^{2} (- (2 x - 1) + 2 (x + 7))}{{(x + 7)}^{4}}$

Step 6.4.2

Apply the distributive property.

$\frac{3 {(2 x - 1)}^{2} (- (2 x) - - 1 + 2 (x + 7))}{{(x + 7)}^{4}}$

Step 6.4.3

Multiply $2$ by $- 1$ .

$\frac{3 {(2 x - 1)}^{2} (- 2 x - - 1 + 2 (x + 7))}{{(x + 7)}^{4}}$

Step 6.4.4

Multiply $- 1$ by $- 1$ .

$\frac{3 {(2 x - 1)}^{2} (- 2 x + 1 + 2 (x + 7))}{{(x + 7)}^{4}}$

Step 6.4.5

Apply the distributive property.

$\frac{3 {(2 x - 1)}^{2} (- 2 x + 1 + 2 x + 2 \cdot 7)}{{(x + 7)}^{4}}$

Step 6.4.6

Multiply $2$ by $7$ .

$\frac{3 {(2 x - 1)}^{2} (- 2 x + 1 + 2 x + 14)}{{(x + 7)}^{4}}$

Step 6.4.7

Add $- 2 x$ and $2 x$ .

$\frac{3 {(2 x - 1)}^{2} (0 + 1 + 14)}{{(x + 7)}^{4}}$

Step 6.4.8

Add $0$ and $1$ .

$\frac{3 {(2 x - 1)}^{2} (1 + 14)}{{(x + 7)}^{4}}$

Step 6.4.9

Add $1$ and $14$ .

$\frac{3 {(2 x - 1)}^{2} \cdot 15}{{(x + 7)}^{4}}$

Step 6.4.10

Multiply $15$ by $3$ .

$\frac{45 {(2 x - 1)}^{2}}{{(x + 7)}^{4}}$