Calculus Examples

f (x) = 5 x + cos (2 x + 1)

$f (x) = 5 x + \cos (2 x + 1)$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of

5 x + cos (2 x + 1)

$5 x + \cos (2 x + 1)$ with respect to

x

$x$ is

\frac{d}{d x} [5 x] + \frac{d}{d x} [cos (2 x + 1)]

$\frac{d}{d x} [5 x] + \frac{d}{d x} [\cos (2 x + 1)]$ .

\frac{d}{d x} [5 x] + \frac{d}{d x} [cos (2 x + 1)]

$\frac{d}{d x} [5 x] + \frac{d}{d x} [\cos (2 x + 1)]$

Step 1.2

Evaluate

\frac{d}{d x} [5 x]

$\frac{d}{d x} [5 x]$ .

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Step 1.2.1

Since

5

$5$ is constant with respect to

x

$x$ , the derivative of

5 x

$5 x$ with respect to

x

$x$ is

5 \frac{d}{d x} [x]

$5 \frac{d}{d x} [x]$ .

5 \frac{d}{d x} [x] + \frac{d}{d x} [cos (2 x + 1)]

$5 \frac{d}{d x} [x] + \frac{d}{d x} [\cos (2 x + 1)]$

Step 1.2.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

5 \cdot 1 + \frac{d}{d x} [cos (2 x + 1)]

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

Step 1.2.3

Multiply

5

$5$ by

1

$1$ .

5 + \frac{d}{d x} [cos (2 x + 1)]

$5 + \frac{d}{d x} [\cos (2 x + 1)]$

5 + \frac{d}{d x} [cos (2 x + 1)]

$5 + \frac{d}{d x} [\cos (2 x + 1)]$

Step 1.3

Evaluate

\frac{d}{d x} [cos (2 x + 1)]

$\frac{d}{d x} [\cos (2 x + 1)]$ .

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Step 1.3.1

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = cos (x)

$f (x) = \cos (x)$ and

g (x) = 2 x + 1

$g (x) = 2 x + 1$ .

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Step 1.3.1.1

To apply the Chain Rule, set

u

$u$ as

2 x + 1

$2 x + 1$ .

5 + \frac{d}{d u} [cos (u)] \frac{d}{d x} [2 x + 1]

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

Step 1.3.1.2

The derivative of

cos (u)

$\cos (u)$ with respect to

u

$u$ is

- sin (u)

$- \sin (u)$ .

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

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

Step 1.3.1.3

Replace all occurrences of

u

$u$ with

2 x + 1

$2 x + 1$ .

5 - sin (2 x + 1) \frac{d}{d x} [2 x + 1]

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

5 - sin (2 x + 1) \frac{d}{d x} [2 x + 1]

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

Step 1.3.2

By the Sum Rule, the derivative of

2 x + 1

$2 x + 1$ with respect to

x

$x$ is

\frac{d}{d x} [2 x] + \frac{d}{d x} [1]

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

5 - sin (2 x + 1) (\frac{d}{d x} [2 x] + \frac{d}{d x} [1])

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

Step 1.3.3

Since

2

$2$ is constant with respect to

x

$x$ , the derivative of

2 x

$2 x$ with respect to

x

$x$ is

2 \frac{d}{d x} [x]

$2 \frac{d}{d x} [x]$ .

5 - sin (2 x + 1) (2 \frac{d}{d x} [x] + \frac{d}{d x} [1])

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

Step 1.3.4

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

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

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

Step 1.3.5

Since

1

$1$ is constant with respect to

x

$x$ , the derivative of

1

$1$ with respect to

x

$x$ is

0

$0$ .

5 - sin (2 x + 1) (2 \cdot 1 + 0)

$5 - \sin (2 x + 1) (2 \cdot 1 + 0)$

Step 1.3.6

Multiply

2

$2$ by

1

$1$ .

5 - sin (2 x + 1) (2 + 0)

$5 - \sin (2 x + 1) (2 + 0)$

Step 1.3.7

Add

2

$2$ and

0

$0$ .

5 - sin (2 x + 1) \cdot 2

$5 - \sin (2 x + 1) \cdot 2$

Step 1.3.8

Multiply

2

$2$ by

- 1

$- 1$ .

5 - 2 sin (2 x + 1)

$5 - 2 \sin (2 x + 1)$

5 - 2 sin (2 x + 1)

$5 - 2 \sin (2 x + 1)$

5 - 2 sin (2 x + 1)

$5 - 2 \sin (2 x + 1)$

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

By the Sum Rule, the derivative of

5 - 2 sin (2 x + 1)

$5 - 2 \sin (2 x + 1)$ with respect to

x

$x$ is

\frac{d}{d x} [5] + \frac{d}{d x} [- 2 sin (2 x + 1)]

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

f'' (x) = \frac{d}{d x} (5) + \frac{d}{d x} (- 2 sin (2 x + 1))

$f'' (x) = \frac{d}{d x} (5) + \frac{d}{d x} (- 2 \sin (2 x + 1))$

Step 2.1.2

Since

5

$5$ is constant with respect to

x

$x$ , the derivative of

5

$5$ with respect to

x

$x$ is

0

$0$ .

f'' (x) = 0 + \frac{d}{d x} (- 2 sin (2 x + 1))

$f'' (x) = 0 + \frac{d}{d x} (- 2 \sin (2 x + 1))$

f'' (x) = 0 + \frac{d}{d x} (- 2 sin (2 x + 1))

$f'' (x) = 0 + \frac{d}{d x} (- 2 \sin (2 x + 1))$

Step 2.2

Evaluate

\frac{d}{d x} [- 2 sin (2 x + 1)]

$\frac{d}{d x} [- 2 \sin (2 x + 1)]$ .

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Step 2.2.1

Since

- 2

$- 2$ is constant with respect to

x

$x$ , the derivative of

- 2 sin (2 x + 1)

$- 2 \sin (2 x + 1)$ with respect to

x

$x$ is

- 2 \frac{d}{d x} [sin (2 x + 1)]

$- 2 \frac{d}{d x} [\sin (2 x + 1)]$ .

f'' (x) = 0 - 2 \frac{d}{d x} sin (2 x + 1)

$f'' (x) = 0 - 2 \frac{d}{d x} \sin (2 x + 1)$

Step 2.2.2

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = sin (x)

$f (x) = \sin (x)$ and

g (x) = 2 x + 1

$g (x) = 2 x + 1$ .

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Step 2.2.2.1

To apply the Chain Rule, set

u

$u$ as

2 x + 1

$2 x + 1$ .

f'' (x) = 0 - 2 (\frac{d}{d u} (sin (u)) \frac{d}{d x} (2 x + 1))

$f'' (x) = 0 - 2 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x + 1))$

Step 2.2.2.2

The derivative of

sin (u)

$\sin (u)$ with respect to

u

$u$ is

cos (u)

$\cos (u)$ .

f'' (x) = 0 - 2 (cos (u) \frac{d}{d x} (2 x + 1))

$f'' (x) = 0 - 2 (\cos (u) \frac{d}{d x} (2 x + 1))$

Step 2.2.2.3

Replace all occurrences of

u

$u$ with

2 x + 1

$2 x + 1$ .

f'' (x) = 0 - 2 (cos (2 x + 1) \frac{d}{d x} (2 x + 1))

$f'' (x) = 0 - 2 (\cos (2 x + 1) \frac{d}{d x} (2 x + 1))$

f'' (x) = 0 - 2 (cos (2 x + 1) \frac{d}{d x} (2 x + 1))

$f'' (x) = 0 - 2 (\cos (2 x + 1) \frac{d}{d x} (2 x + 1))$

Step 2.2.3

By the Sum Rule, the derivative of

2 x + 1

$2 x + 1$ with respect to

x

$x$ is

\frac{d}{d x} [2 x] + \frac{d}{d x} [1]

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

f'' (x) = 0 - 2 (cos (2 x + 1) (\frac{d}{d x} (2 x) + \frac{d}{d x} (1)))

$f'' (x) = 0 - 2 (\cos (2 x + 1) (\frac{d}{d x} (2 x) + \frac{d}{d x} (1)))$

Step 2.2.4

Since

2

$2$ is constant with respect to

x

$x$ , the derivative of

2 x

$2 x$ with respect to

x

$x$ is

2 \frac{d}{d x} [x]

$2 \frac{d}{d x} [x]$ .

f'' (x) = 0 - 2 (cos (2 x + 1) (2 \frac{d}{d x} (x) + \frac{d}{d x} (1)))

$f'' (x) = 0 - 2 (\cos (2 x + 1) (2 \frac{d}{d x} (x) + \frac{d}{d x} (1)))$

Step 2.2.5

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

f'' (x) = 0 - 2 (cos (2 x + 1) (2 \cdot 1 + \frac{d}{d x} (1)))

$f'' (x) = 0 - 2 (\cos (2 x + 1) (2 \cdot 1 + \frac{d}{d x} (1)))$

Step 2.2.6

Since

1

$1$ is constant with respect to

x

$x$ , the derivative of

1

$1$ with respect to

x

$x$ is

0

$0$ .

f'' (x) = 0 - 2 (cos (2 x + 1) (2 \cdot 1 + 0))

$f'' (x) = 0 - 2 (\cos (2 x + 1) (2 \cdot 1 + 0))$

Step 2.2.7

Multiply

2

$2$ by

1

$1$ .

f'' (x) = 0 - 2 (cos (2 x + 1) (2 + 0))

$f'' (x) = 0 - 2 (\cos (2 x + 1) (2 + 0))$

Step 2.2.8

Add

2

$2$ and

0

$0$ .

f'' (x) = 0 - 2 (cos (2 x + 1) \cdot 2)

$f'' (x) = 0 - 2 (\cos (2 x + 1) \cdot 2)$

Step 2.2.9

Move

2

$2$ to the left of

cos (2 x + 1)

$\cos (2 x + 1)$ .

f'' (x) = 0 - 2 (2 \cdot cos (2 x + 1))

$f'' (x) = 0 - 2 (2 \cdot \cos (2 x + 1))$

Step 2.2.10

Multiply

2

$2$ by

- 2

$- 2$ .

f'' (x) = 0 - 4 cos (2 x + 1)

$f'' (x) = 0 - 4 \cos (2 x + 1)$

f'' (x) = 0 - 4 cos (2 x + 1)

$f'' (x) = 0 - 4 \cos (2 x + 1)$

Step 2.3

Subtract

4 cos (2 x + 1)

$4 \cos (2 x + 1)$ from

0

$0$ .

f'' (x) = - 4 cos (2 x + 1)

$f'' (x) = - 4 \cos (2 x + 1)$

f'' (x) = - 4 cos (2 x + 1)

$f'' (x) = - 4 \cos (2 x + 1)$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

0

$0$ and solve.

5 - 2 sin (2 x + 1) = 0

$5 - 2 \sin (2 x + 1) = 0$

Step 4

Subtract

5

$5$ from both sides of the equation.

- 2 sin (2 x + 1) = - 5

$- 2 \sin (2 x + 1) = - 5$

Step 5

Divide each term in

- 2 sin (2 x + 1) = - 5

$- 2 \sin (2 x + 1) = - 5$ by

- 2

$- 2$ and simplify.

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Step 5.1

Divide each term in

- 2 sin (2 x + 1) = - 5

$- 2 \sin (2 x + 1) = - 5$ by

- 2

$- 2$ .

\frac{- 2 sin (2 x + 1)}{- 2} = \frac{- 5}{- 2}

$\frac{- 2 \sin (2 x + 1)}{- 2} = \frac{- 5}{- 2}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of

- 2

$- 2$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{- 2 \sin (2 x + 1)}{- 2} = \frac{- 5}{- 2}$

Step 5.2.1.2

Divide

$\sin (2 x + 1)$ by

$1$ .

$\sin (2 x + 1) = \frac{- 5}{- 2}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Dividing two negative values results in a positive value.

$\sin (2 x + 1) = \frac{5}{2}$

Step 6

The range of sine is

$- 1 \leq y \leq 1$ . Since

$\frac{5}{2}$ does not fall in this range, there is no solution.

No solution

Step 7

Evaluate the second derivative at

$x =$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 \cos (2 + 1)$

Step 8

Evaluate the second derivative.

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Step 8.1

Add

$2$ and

$1$ .

$- 4 \cos (3)$

Step 8.2

Evaluate

$\cos (3)$ .

$- 4 \cdot 0.99862953$

Step 8.3

Multiply

$- 4$ by

$0.99862953$ .

$- 3.99451813$

Step 9

$x =$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x =$ is a local maximum

Step 10

These are the local extrema for

$f (x) = 5 x + \cos (2 x + 1)$ .

$(, i s a (l o c a l) (m a x i m u m))$ is a local maxima

Step 11