Calculus Examples

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

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

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

Step 1.1.2

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

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

Step 1.2

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

$1 + \cos (x)$

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

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

Step 2.1.2

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

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

$1 + \cos (x) = 0$

Step 4

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

Step 5

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

$x = \arccos (- 1)$

Step 6

Simplify the right side.

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Step 6.1

The exact value of $\arccos (- 1)$ is $π$ .

$x = π$

Step 7

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Step 8

Subtract $π$ from $2 π$ .

$x = π$

Step 9

The solution to the equation $x = π$ .

$x = π, π$

Step 10

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

$- \sin (π)$

Step 11

Evaluate the second derivative.

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Step 11.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$- \sin (0)$

Step 11.2

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

$- 0$

Step 11.3

Multiply $- 1$ by $0$ .

$0$

Step 12

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

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Step 12.1

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

$(- \infty, π) \cup (π, \infty)$

Step 12.2

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

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Step 12.2.1

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

$f' (0) = 1 + \cos (0)$

Step 12.2.2

Simplify the result.

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Step 12.2.2.1

The exact value of $\cos (0)$ is $1$ .

$f' (0) = 1 + 1$

Step 12.2.2.2

Add $1$ and $1$ .

$f' (0) = 2$

Step 12.2.2.3

The final answer is $2$ .

$2$

Step 12.3

Substitute any number, such as $6$ , from the interval $(π, \infty)$ in the first derivative $1 + \cos (x)$ to check if the result is negative or positive.

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Step 12.3.1

Replace the variable $x$ with $6$ in the expression.

$f' (6) = 1 + \cos (6)$

Step 12.3.2

Simplify the result.

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Step 12.3.2.1

Evaluate $\cos (6)$ .

$f' (6) = 1 + 0.96017028$

Step 12.3.2.2

Add $1$ and $0.96017028$ .

$f' (6) = 1.96017028$

Step 12.3.2.3

The final answer is $1.96017028$ .

$1.96017028$

Step 12.4

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

Not a local maximum or minimum

Step 12.5

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

No local maxima or minima

Step 13