Calculus Examples

$f (x) = \sin (\frac{x}{2})$

Step 1

Find the second derivative.

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Find the first derivative.

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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 $g (x) = \frac{x}{2}$ .

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To apply the Chain Rule, set $u$ as $\frac{x}{2}$ .

$f' (x) = \frac{d}{d u} (\sin (u)) \frac{d}{d x} (\frac{x}{2})$

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

$f' (x) = \cos (u) \frac{d}{d x} (\frac{x}{2})$

Replace all occurrences of $u$ with $\frac{x}{2}$ .

$f' (x) = \cos (\frac{x}{2}) \frac{d}{d x} (\frac{x}{2})$

Differentiate.

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

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

Combine $\frac{1}{2}$ and $\cos (\frac{x}{2})$ .

$f' (x) = \frac{\cos (\frac{x}{2})}{2} \cdot \frac{d}{d x} (x)$

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

$f' (x) = \frac{\cos (\frac{x}{2})}{2} \cdot 1$

Multiply $\frac{\cos (\frac{x}{2})}{2}$ by $1$ .

$f' (x) = \frac{\cos (\frac{x}{2})}{2}$

Find the second derivative.

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

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

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

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To apply the Chain Rule, set $u$ as $\frac{x}{2}$ .

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

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

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

Replace all occurrences of $u$ with $\frac{x}{2}$ .

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

Differentiate.

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Combine $\frac{1}{2}$ and $\sin (\frac{x}{2})$ .

$f'' (x) = - \frac{\sin (\frac{x}{2})}{2} \cdot \frac{d}{d x} \frac{x}{2}$

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

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

Combine fractions.

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Multiply $\frac{1}{2}$ by $\frac{\sin (\frac{x}{2})}{2}$ .

$f'' (x) = - \frac{\sin (\frac{x}{2})}{2 \cdot 2} \cdot \frac{d}{d x} x$

Multiply $2$ by $2$ .

$f'' (x) = - \frac{\sin (\frac{x}{2})}{4} \cdot \frac{d}{d x} x$

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

$f'' (x) = - \frac{\sin (\frac{x}{2})}{4} \cdot 1$

Multiply $- 1$ by $1$ .

$f'' (x) = - \frac{\sin (\frac{x}{2})}{4}$

The second derivative of $f (x)$ with respect to $x$ is $- \frac{\sin (\frac{x}{2})}{4}$ .

$- \frac{\sin (\frac{x}{2})}{4}$

Step 2

Set the second derivative equal to $0$ then solve the equation $- \frac{\sin (\frac{x}{2})}{4} = 0$ .

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Set the second derivative equal to $0$ .

$- \frac{\sin (\frac{x}{2})}{4} = 0$

Set the numerator equal to zero.

$\sin (\frac{x}{2}) = 0$

Solve the equation for $x$ .

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

$\frac{x}{2} = \arcsin (0)$

Simplify the right side.

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

$\frac{x}{2} = 0$

Set the numerator equal to zero.

$x = 0$

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.

$\frac{x}{2} = π - 0$

Solve for $x$ .

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Multiply both sides of the equation by $2$ .

$2 \frac{x}{2} = 2 (π - 0)$

Simplify both sides of the equation.

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Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$2 \frac{x}{2} = 2 (π - 0)$

Rewrite the expression.

$x = 2 (π - 0)$

Simplify the right side.

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Subtract $0$ from $π$ .

$x = 2 π$

Find the period of $\sin (\frac{x}{2})$ .

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The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $\frac{1}{2}$ in the formula for period.

$\frac{2 π}{| \frac{1}{2} |}$

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Multiply $2$ by $2$ .

$4 π$

The period of the $\sin (\frac{x}{2})$ function is $4 π$ so values will repeat every $4 π$ radians in both directions.

$x = 4 π n, 2 π + 4 π n$ , for any integer $n$

Consolidate the answers.

$x = 2 π n$ , for any integer $n$

Step 3

The point found by substituting in $f (x) = \sin (\frac{x}{2})$ is $(,)$ . This point can be an inflection point.

$(,)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty,) \cup (, \infty)$

Step 5

Substitute a value from the interval $(- \infty,)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = - \frac{\sin (\frac{- 0.1}{2})}{4}$

Simplify the result.

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Simplify the numerator.

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Divide $- 0.1$ by $2$ .

$f'' (- 0.1) = - \frac{\sin (- 0.05)}{4}$

Evaluate $\sin (- 0.05)$ .

$f'' (- 0.1) = - \frac{- 0.04997916}{4}$

Simplify the expression.

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Divide $- 0.04997916$ by $4$ .

$f'' (- 0.1) = 0.01249479$

Multiply $- 1$ by $- 0.01249479$ .

$f'' (- 0.1) = 0.01249479$

The final answer is $0.01249479$ .

$0.01249479$

At $- 0.1$ , the second derivative is $0.01249479$ . Since this is positive, the second derivative is increasing on the interval $(- \infty,)$ .

Increasing on $(- \infty,)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = - \frac{\sin (\frac{0.1}{2})}{4}$

Simplify the result.

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Simplify the numerator.

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Divide $0.1$ by $2$ .

$f'' (0.1) = - \frac{\sin (0.05)}{4}$

Evaluate $\sin (0.05)$ .

$f'' (0.1) = - \frac{0.04997916}{4}$

Simplify the expression.

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Divide $0.04997916$ by $4$ .

$f'' (0.1) = - 1 \cdot 0.01249479$

Multiply $- 1$ by $0.01249479$ .

$f'' (0.1) = - 0.01249479$

The final answer is $- 0.01249479$ .

$- 0.01249479$

At $0.1$ , the second derivative is $- 0.01249479$ . Since this is negative, the second derivative is decreasing on the interval $(, \infty)$

Decreasing on $(, \infty)$ since $f'' (x) < 0$

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(,)$ .

$(,)$

Step 8