Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $1 {(x - 5)}^{5} + (x - 1) (5 {(x - 1)}^{4} (1))$ with respect to $x$ is $\frac{d}{d x} [1 {(x - 5)}^{5}] + \frac{d}{d x} [(x - 1) (5 {(x - 1)}^{4} (1))]$ .

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

Step 1.2

Evaluate $\frac{d}{d x} [1 {(x - 5)}^{5}]$ .

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Step 1.2.1

Multiply ${(x - 5)}^{5}$ by $1$ .

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

Step 1.2.2

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

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Step 1.2.2.1

To apply the Chain Rule, set $u_{1}$ as $x - 5$ .

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

Step 1.2.2.2

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

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

Step 1.2.2.3

Replace all occurrences of $u_{1}$ with $x - 5$ .

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

Add $1$ and $0$ .

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

Step 1.2.7

Multiply $5$ by $1$ .

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

Step 1.3

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

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Step 1.3.1

Multiply $5$ by $1$ .

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

Step 1.3.2

Raise $x - 1$ to the power of $1$ .

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

Step 1.3.3

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

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

Step 1.3.4

Add $1$ and $4$ .

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

Step 1.3.5

Since $5$ is constant with respect to $x$ , the derivative of $5 {(x - 1)}^{5}$ with respect to $x$ is $5 \frac{d}{d x} [{(x - 1)}^{5}]$ .

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

Step 1.3.6

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

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Step 1.3.6.1

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

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

Step 1.3.6.2

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

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

Step 1.3.6.3

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

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

Step 1.3.7

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

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

Step 1.3.8

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

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

Step 1.3.9

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

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

Step 1.3.10

Add $1$ and $0$ .

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

Step 1.3.11

Multiply $5$ by $1$ .

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

Step 1.3.12

Multiply $5$ by $5$ .

$5 {(x - 5)}^{4} + 25 {(x - 1)}^{4}$

Step 1.4

Factor $5$ out of $5 {(x - 5)}^{4} + 25 {(x - 1)}^{4}$ .

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Step 1.4.1

Factor $5$ out of $5 {(x - 5)}^{4}$ .

$5 ({(x - 5)}^{4}) + 25 {(x - 1)}^{4}$

Step 1.4.2

Factor $5$ out of $25 {(x - 1)}^{4}$ .

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

Step 1.4.3

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

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

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

Since $5$ is constant with respect to $x$ , the derivative of $5 ({(x - 5)}^{4} + 5 {(x - 1)}^{4})$ with respect to $x$ is $5 \frac{d}{d x} [{(x - 5)}^{4} + 5 {(x - 1)}^{4}]$ .

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

Step 2.1.2

By the Sum Rule, the derivative of ${(x - 5)}^{4} + 5 {(x - 1)}^{4}$ with respect to $x$ is $\frac{d}{d x} [{(x - 5)}^{4}] + \frac{d}{d x} [5 {(x - 1)}^{4}]$ .

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

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 $f (x) = x^{4}$ and $g (x) = x - 5$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $x - 5$ .

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

Step 2.2.2

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

$f'' (x) = 5 (4 {u_{1}}^{3} \frac{d}{d x} (x - 5) + \frac{d}{d x} (5 {(x - 1)}^{4}))$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $x - 5$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} \frac{d}{d x} (x - 5) + \frac{d}{d x} (5 {(x - 1)}^{4}))$

Step 2.3

Differentiate.

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Step 2.3.1

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

$f'' (x) = 5 (4 {(x - 5)}^{3} (\frac{d}{d x} (x) + \frac{d}{d x} (- 5)) + \frac{d}{d x} (5 {(x - 1)}^{4}))$

Step 2.3.2

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

$f'' (x) = 5 (4 {(x - 5)}^{3} (1 + \frac{d}{d x} (- 5)) + \frac{d}{d x} (5 {(x - 1)}^{4}))$

Step 2.3.3

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

$f'' (x) = 5 (4 {(x - 5)}^{3} (1 + 0) + \frac{d}{d x} (5 {(x - 1)}^{4}))$

Step 2.3.4

Simplify the expression.

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Step 2.3.4.1

Add $1$ and $0$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} \cdot 1 + \frac{d}{d x} (5 {(x - 1)}^{4}))$

Step 2.3.4.2

Multiply $4$ by $1$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} + \frac{d}{d x} (5 {(x - 1)}^{4}))$

Step 2.3.5

Since $5$ is constant with respect to $x$ , the derivative of $5 {(x - 1)}^{4}$ with respect to $x$ is $5 \frac{d}{d x} [{(x - 1)}^{4}]$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} + 5 \frac{d}{d x} ({(x - 1)}^{4}))$

Step 2.4

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

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Step 2.4.1

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

$f'' (x) = 5 (4 {(x - 5)}^{3} + 5 (\frac{d}{d u (2)} ({u_{2}}^{4}) \frac{d}{d x} (x - 1)))$

Step 2.4.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 4$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} + 5 (4 {u_{2}}^{3} \frac{d}{d x} (x - 1)))$

Step 2.4.3

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

$f'' (x) = 5 (4 {(x - 5)}^{3} + 5 (4 {(x - 1)}^{3} \frac{d}{d x} (x - 1)))$

Step 2.5

Differentiate.

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Step 2.5.1

Multiply $4$ by $5$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} + 20 ({(x - 1)}^{3} \frac{d}{d x} (x - 1)))$

Step 2.5.2

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

$f'' (x) = 5 (4 {(x - 5)}^{3} + 20 {(x - 1)}^{3} (\frac{d}{d x} (x) + \frac{d}{d x} (- 1)))$

Step 2.5.3

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

$f'' (x) = 5 (4 {(x - 5)}^{3} + 20 {(x - 1)}^{3} (1 + \frac{d}{d x} (- 1)))$

Step 2.5.4

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

$f'' (x) = 5 (4 {(x - 5)}^{3} + 20 {(x - 1)}^{3} (1 + 0))$

Step 2.5.5

Simplify the expression.

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Step 2.5.5.1

Add $1$ and $0$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} + 20 {(x - 1)}^{3} \cdot 1)$

Step 2.5.5.2

Multiply $20$ by $1$ .

$f'' (x) = 5 (4 {(x - 5)}^{3} + 20 {(x - 1)}^{3})$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$f'' (x) = 5 (4 {(x - 5)}^{3}) + 5 (20 {(x - 1)}^{3})$

Step 2.6.2

Multiply $4$ by $5$ .

$f'' (x) = 20 {(x - 5)}^{3} + 5 (20 {(x - 1)}^{3})$

Step 2.6.3

Multiply $20$ by $5$ .

$f'' (x) = 20 {(x - 5)}^{3} + 100 {(x - 1)}^{3}$

Step 2.6.4

Factor $20$ out of $20 {(x - 5)}^{3} + 100 {(x - 1)}^{3}$ .

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Step 2.6.4.1

Factor $20$ out of $20 {(x - 5)}^{3}$ .

$f'' (x) = 20 ({(x - 5)}^{3}) + 100 {(x - 1)}^{3}$

Step 2.6.4.2

Factor $20$ out of $100 {(x - 1)}^{3}$ .

$f'' (x) = 20 ({(x - 5)}^{3}) + 20 (5 {(x - 1)}^{3})$

Step 2.6.4.3

Factor $20$ out of $20 ({(x - 5)}^{3}) + 20 (5 {(x - 1)}^{3})$ .

$f'' (x) = 20 ({(x - 5)}^{3} + 5 {(x - 1)}^{3})$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

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

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema

Step 6