Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $5 x + \cos (2 x + 1)$ with respect to $x$ is $\frac{d}{d x} [5 x] + \frac{d}{d x} [\cos (2 x + 1)]$ .

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

Multiply $5$ by $1$ .

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

Step 1.3

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

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Step 1.3.1

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

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Step 1.3.1.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

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

Step 1.3.1.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

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

Step 1.3.1.3

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

Multiply $2$ by $1$ .

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

Step 1.3.7

Add $2$ and $0$ .

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

Step 1.3.8

Multiply $2$ by $- 1$ .

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

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

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \sin (2 x + 1)$ with respect to $x$ is $- 2 \frac{d}{d x} [\sin (2 x + 1)]$ .

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

Step 2.2.2

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

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

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

Step 2.2.2.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

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

Step 2.2.2.3

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

Multiply $2$ by $1$ .

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

Step 2.2.8

Add $2$ and $0$ .

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

Step 2.2.9

Move $2$ to the left of $\cos (2 x + 1)$ .

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

Step 2.2.10

Multiply $2$ by $- 2$ .

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

Step 2.3

Subtract $4 \cos (2 x + 1)$ from $0$ .

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

Step 3

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

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

Step 4

Subtract $5$ from both sides of the equation.

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

Step 5

Divide each term in $- 2 \sin (2 x + 1) = - 5$ by $- 2$ and simplify.

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Step 5.1

Divide each term in $- 2 \sin (2 x + 1) = - 5$ by $- 2$ .

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $- 2$ .

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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