Calculus Examples

$f (x) = 4 \sin (\frac{x}{2}) - 3$

Find the first derivative of the function.

Tap for more steps...

By the Sum Rule, the derivative of $4 \sin (\frac{x}{2}) - 3$ with respect to $x$ is $\frac{d}{d x} [4 \sin (\frac{x}{2})] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [4 \sin (\frac{x}{2})] + \frac{d}{d x} [- 3]$

Evaluate $\frac{d}{d x} [4 \sin (\frac{x}{2})]$ .

Tap for more steps...

Since $4$ is constant with respect to $x$ , the derivative of $4 \sin (\frac{x}{2})$ with respect to $x$ is $4 \frac{d}{d x} [\sin (\frac{x}{2})]$ .

$4 \frac{d}{d x} [\sin (\frac{x}{2})] + \frac{d}{d x} [- 3]$

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

Tap for more steps...

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

$4 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [\frac{x}{2}]) + \frac{d}{d x} [- 3]$

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

$4 (\cos (u) \frac{d}{d x} [\frac{x}{2}]) + \frac{d}{d x} [- 3]$

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

$4 (\cos (\frac{x}{2}) \frac{d}{d x} [\frac{x}{2}]) + \frac{d}{d x} [- 3]$

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

$4 (\cos (\frac{x}{2}) (\frac{1}{2} \frac{d}{d x} [x])) + \frac{d}{d x} [- 3]$

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

$4 (\cos (\frac{x}{2}) (\frac{1}{2} \cdot 1)) + \frac{d}{d x} [- 3]$

Multiply $\frac{1}{2}$ by $1$ .

$4 (\cos (\frac{x}{2}) \frac{1}{2}) + \frac{d}{d x} [- 3]$

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

$4 \frac{\cos (\frac{x}{2})}{2} + \frac{d}{d x} [- 3]$

Combine $4$ and $\frac{\cos (\frac{x}{2})}{2}$ .

$\frac{4 \cos (\frac{x}{2})}{2} + \frac{d}{d x} [- 3]$

Cancel the common factor of $4$ and $2$ .

Tap for more steps...

Factor $2$ out of $4 \cos (\frac{x}{2})$ .

$\frac{2 (2 \cos (\frac{x}{2}))}{2} + \frac{d}{d x} [- 3]$

Cancel the common factors.

Tap for more steps...

Factor $2$ out of $2$ .

$\frac{2 (2 \cos (\frac{x}{2}))}{2 (1)} + \frac{d}{d x} [- 3]$

Cancel the common factor.

$\frac{2 (2 \cos (\frac{x}{2}))}{2 \cdot 1} + \frac{d}{d x} [- 3]$

Rewrite the expression.

$\frac{2 \cos (\frac{x}{2})}{1} + \frac{d}{d x} [- 3]$

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

$2 \cos (\frac{x}{2}) + \frac{d}{d x} [- 3]$

Differentiate using the Constant Rule.

Tap for more steps...

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$2 \cos (\frac{x}{2}) + 0$

Add $2 \cos (\frac{x}{2})$ and $0$ .

$2 \cos (\frac{x}{2})$

Find the second derivative of the function.

Tap for more steps...

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

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

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

Tap for more steps...

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

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

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

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

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

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

Differentiate.

Tap for more steps...

Multiply $- 1$ by $2$ .

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

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

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

Simplify terms.

Tap for more steps...

Combine $\frac{1}{2}$ and $- 2$ .

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

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

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

Cancel the common factor of $- 2$ and $2$ .

Tap for more steps...

Factor $2$ out of $- 2 \sin (\frac{x}{2})$ .

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

Cancel the common factors.

Tap for more steps...

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

Multiply $- 1$ by $1$ .

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

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

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

Divide each term in $2 \cos (\frac{x}{2}) = 0$ by $2$ and simplify.

Tap for more steps...

Divide each term in $2 \cos (\frac{x}{2}) = 0$ by $2$ .

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

Simplify the left side.

Tap for more steps...

Cancel the common factor of $2$ .

Tap for more steps...

Cancel the common factor.

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

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

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

Simplify the right side.

Tap for more steps...

Divide $0$ by $2$ .

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

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

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

Simplify the right side.

Tap for more steps...

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

$\frac{x}{2} = \frac{π}{2}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = π$

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

$\frac{x}{2} = 2 π - \frac{π}{2}$

Solve for $x$ .

Tap for more steps...

Multiply both sides of the equation by $2$ .

$2 \frac{x}{2} = 2 (2 π - \frac{π}{2})$

Simplify both sides of the equation.

Tap for more steps...

Simplify the left side.

Tap for more steps...

Cancel the common factor of $2$ .

Tap for more steps...

Cancel the common factor.

$2 \frac{x}{2} = 2 (2 π - \frac{π}{2})$

Rewrite the expression.

$x = 2 (2 π - \frac{π}{2})$

Simplify the right side.

Tap for more steps...

Simplify $2 (2 π - \frac{π}{2})$ .

Tap for more steps...

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

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

Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$x = 2 \frac{2 π \cdot 2 - π}{2}$

Cancel the common factor of $2$ .

Tap for more steps...

Cancel the common factor.

$x = 2 \frac{2 π \cdot 2 - π}{2}$

Rewrite the expression.

$x = 2 π \cdot 2 - π$

Multiply $2$ by $2$ .

$x = 4 π - π$

Subtract $π$ from $4 π$ .

$x = 3 π$

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

$x = π, 3 π$

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.

$- \sin (\frac{π}{2})$

Evaluate the second derivative.

Tap for more steps...

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

$- 1 \cdot 1$

Multiply $- 1$ by $1$ .

$- 1$

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

$x = π$ is a local maximum

Find the y-value when $x = π$ .

Tap for more steps...

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

$f (π) = 4 \sin (\frac{π}{2}) - 3$

Simplify the result.

Tap for more steps...

Simplify each term.

Tap for more steps...

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

$f (π) = 4 \cdot 1 - 3$

Multiply $4$ by $1$ .

$f (π) = 4 - 3$

Subtract $3$ from $4$ .

$f (π) = 1$

The final answer is $1$ .

$y = 1$

Evaluate the second derivative at $x = 3 π$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin (\frac{3 π}{2})$

Evaluate the second derivative.

Tap for more steps...

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$- - \sin (\frac{π}{2})$

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

$- (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$- - 1$

Multiply $- 1$ by $- 1$ .

$1$

$x = 3 π$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 3 π$ is a local minimum

Find the y-value when $x = 3 π$ .

Tap for more steps...

Replace the variable $x$ with $3 π$ in the expression.

$f (3 π) = 4 \sin (\frac{3 π}{2}) - 3$

Simplify the result.

Tap for more steps...

Simplify each term.

Tap for more steps...

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$f (3 π) = 4 (- \sin (\frac{π}{2})) - 3$

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

$f (3 π) = 4 (- 1 \cdot 1) - 3$

Multiply $- 1$ by $1$ .

$f (3 π) = 4 \cdot - 1 - 3$

Multiply $4$ by $- 1$ .

$f (3 π) = - 4 - 3$

Subtract $3$ from $- 4$ .

$f (3 π) = - 7$

The final answer is $- 7$ .

$y = - 7$

These are the local extrema for $f (x) = 4 \sin (\frac{x}{2}) - 3$ .

$(π, 1)$ is a local maxima

$(3 π, - 7)$ is a local minima