Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.2

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

$1 - \sin (x)$

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

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

$f'' (x) = 0 - \cos (x)$

Step 2.3

Subtract $\cos (x)$ from $0$ .

$f'' (x) = - \cos (x)$

Step 3

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

$1 - \sin (x) = 0$

Step 4

Subtract $1$ from both sides of the equation.

$- \sin (x) = - 1$

Step 5

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

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Step 5.1

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

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2

Divide $\sin (x)$ by $1$ .

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Divide $- 1$ by $- 1$ .

$\sin (x) = 1$

Step 6

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (1)$

Step 7

Simplify the right side.

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Step 7.1

The exact value of $\arcsin (1)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 8

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{2}$

Step 9

Simplify $π - \frac{π}{2}$ .

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Step 9.1

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

$x = π \cdot \frac{2}{2} - \frac{π}{2}$

Step 9.2

Combine fractions.

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Step 9.2.1

Combine $π$ and $\frac{2}{2}$ .

$x = \frac{π \cdot 2}{2} - \frac{π}{2}$

Step 9.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 9.3

Simplify the numerator.

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Step 9.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 9.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 10

The solution to the equation $x = \frac{π}{2}$ .

$x = \frac{π}{2}, \frac{π}{2}$

Step 11

Evaluate the second derivative at $x = \frac{π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \cos (\frac{π}{2})$

Step 12

Evaluate the second derivative.

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Step 12.1

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$- 0$

Step 12.2

Multiply $- 1$ by $0$ .

$0$

Step 13

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 13.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, \frac{π}{2}) \cup (\frac{π}{2}, \infty)$

Step 13.2

Substitute any number, such as $0$ , from the interval $(- \infty, \frac{π}{2})$ in the first derivative $1 - \sin (x)$ to check if the result is negative or positive.

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Step 13.2.1

Replace the variable $x$ with $0$ in the expression.

$f' (0) = 1 - \sin (0)$

Step 13.2.2

Simplify the result.

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Step 13.2.2.1

Simplify each term.

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Step 13.2.2.1.1

The exact value of $\sin (0)$ is $0$ .

$f' (0) = 1 - 0$

Step 13.2.2.1.2

Multiply $- 1$ by $0$ .

$f' (0) = 1 + 0$

Step 13.2.2.2

Add $1$ and $0$ .

$f' (0) = 1$

Step 13.2.2.3

The final answer is $1$ .

$1$

Step 13.3

Substitute any number, such as $4$ , from the interval $(\frac{π}{2}, \infty)$ in the first derivative $1 - \sin (x)$ to check if the result is negative or positive.

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Step 13.3.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = 1 - \sin (4)$

Step 13.3.2

Simplify the result.

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Step 13.3.2.1

Simplify each term.

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Step 13.3.2.1.1

Evaluate $\sin (4)$ .

$f' (4) = 1 + 0.75680249$

Step 13.3.2.1.2

Multiply $- 1$ by $- 0.75680249$ .

$f' (4) = 1 + 0.75680249$

Step 13.3.2.2

Add $1$ and $0.75680249$ .

$f' (4) = 1.75680249$

Step 13.3.2.3

The final answer is $1.75680249$ .

$1.75680249$

Step 13.4

Since the first derivative did not change signs around $x = \frac{π}{2}$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 13.5

No local maxima or minima found for $f (x) = x + \cos (x)$ .

No local maxima or minima

Step 14