Calculus Examples

$f (x) = 2 \cos (x) + \sin^{2} (x)$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

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

$\frac{d}{d x} [2 \cos (x)] + \frac{d}{d x} [\sin^{2} (x)]$

Step 1.2

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

Tap for more steps...

Step 1.2.1

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

$2 \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [\sin^{2} (x)]$

Step 1.2.2

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

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

Step 1.2.3

Multiply $- 1$ by $2$ .

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

Step 1.3

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

Tap for more steps...

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 $f (x) = x^{2}$ and $g (x) = \sin (x)$ .

Tap for more steps...

Step 1.3.1.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

Step 1.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

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

Step 1.3.1.3

Replace all occurrences of $u$ with $\sin (x)$ .

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

Step 1.3.2

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

$- 2 \sin (x) + 2 \sin (x) \cos (x)$

Step 1.4

Simplify.

Tap for more steps...

Step 1.4.1

Reorder terms.

$2 \cos (x) \sin (x) - 2 \sin (x)$

Step 1.4.2

Simplify each term.

Tap for more steps...

Step 1.4.2.1

Reorder $2 \cos (x)$ and $\sin (x)$ .

$\sin (x) (2 \cos (x)) - 2 \sin (x)$

Step 1.4.2.2

Reorder $\sin (x)$ and $2$ .

$2 \cdot \sin (x) \cos (x) - 2 \sin (x)$

Step 1.4.2.3

Apply the sine double-angle identity.

$\sin (2 x) - 2 \sin (x)$

Step 2

Find the second derivative of the function.

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

Tap for more steps...

Step 2.2.1.1

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $1$ .

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

Step 2.2.5

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

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

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

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

Step 2.3.2

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

$f'' (x) = 2 \cos (2 x) - 2 \cos (x)$

Step 3

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

$\sin (2 x) - 2 \sin (x) = 0$

Step 4

Apply the sine double-angle identity.

$2 \sin (x) \cos (x) - 2 \sin (x) = 0$

Step 5

Factor $2 \sin (x)$ out of $2 \sin (x) \cos (x) - 2 \sin (x)$ .

Tap for more steps...

Step 5.1

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

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

Step 5.2

Factor $2 \sin (x)$ out of $2 \sin (x) \cos (x) + 2 \sin (x) \cdot - 1$ .

$2 \sin (x) (\cos (x) - 1) = 0$

Step 6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sin (x) = 0$

$\cos (x) - 1 = 0$

Step 7

Set $\sin (x)$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 7.1

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Step 7.2

Solve $\sin (x) = 0$ for $x$ .

Tap for more steps...

Step 7.2.1

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

$x = \arcsin (0)$

Step 7.2.2

Simplify the right side.

Tap for more steps...

Step 7.2.2.1

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

$x = 0$

Step 7.2.3

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.

$x = π - 0$

Step 7.2.4

Subtract $0$ from $π$ .

$x = π$

Step 7.2.5

The solution to the equation $x = 0$ .

$x = 0, π$

Step 8

Set $\cos (x) - 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 8.1

Set $\cos (x) - 1$ equal to $0$ .

$\cos (x) - 1 = 0$

Step 8.2

Solve $\cos (x) - 1 = 0$ for $x$ .

Tap for more steps...

Step 8.2.1

Add $1$ to both sides of the equation.

$\cos (x) = 1$

Step 8.2.2

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

$x = \arccos (1)$

Step 8.2.3

Simplify the right side.

Tap for more steps...

Step 8.2.3.1

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

$x = 0$

Step 8.2.4

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.

$x = 2 π - 0$

Step 8.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 8.2.6

The solution to the equation $x = 0$ .

$x = 0, 2 π$

Step 9

The final solution is all the values that make $2 \sin (x) (\cos (x) - 1) = 0$ true.

$x = 0, π, 2 π$

Step 10

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \cos (2 (0)) - 2 \cos (0)$

Step 11

Evaluate the second derivative.

Tap for more steps...

Step 11.1

Simplify each term.

Tap for more steps...

Step 11.1.1

Multiply $2$ by $0$ .

$2 \cos (0) - 2 \cos (0)$

Step 11.1.2

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

$2 \cdot 1 - 2 \cos (0)$

Step 11.1.3

Multiply $2$ by $1$ .

$2 - 2 \cos (0)$

Step 11.1.4

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

$2 - 2 \cdot 1$

Step 11.1.5

Multiply $- 2$ by $1$ .

$2 - 2$

Step 11.2

Subtract $2$ from $2$ .

$0$

Step 12

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

Tap for more steps...

Step 12.1

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

$(- \infty, 0) \cup (0, π) \cup (π, 2 π) \cup (2 π, \infty)$

Step 12.2

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

Tap for more steps...

Step 12.2.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = \sin (2 (- 2)) - 2 \sin (- 2)$

Step 12.2.2

Simplify the result.

Tap for more steps...

Step 12.2.2.1

Simplify each term.

Tap for more steps...

Step 12.2.2.1.1

Multiply $2$ by $- 2$ .

$f' (- 2) = \sin (- 4) - 2 \sin (- 2)$

Step 12.2.2.1.2

Evaluate $\sin (- 4)$ .

$f' (- 2) = 0.75680249 - 2 \sin (- 2)$

Step 12.2.2.1.3

Evaluate $\sin (- 2)$ .

$f' (- 2) = 0.75680249 - 2 \cdot - 0.90929742$

Step 12.2.2.1.4

Multiply $- 2$ by $- 0.90929742$ .

$f' (- 2) = 0.75680249 + 1.81859485$

Step 12.2.2.2

Add $0.75680249$ and $1.81859485$ .

$f' (- 2) = 2.57539734$

Step 12.2.2.3

The final answer is $2.57539734$ .

$2.57539734$

Step 12.3

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

Tap for more steps...

Step 12.3.1

Replace the variable $x$ with $2$ in the expression.

$f' (2) = \sin (2 (2)) - 2 \sin (2)$

Step 12.3.2

Simplify the result.

Tap for more steps...

Step 12.3.2.1

Simplify each term.

Tap for more steps...

Step 12.3.2.1.1

Multiply $2$ by $2$ .

$f' (2) = \sin (4) - 2 \sin (2)$

Step 12.3.2.1.2

Evaluate $\sin (4)$ .

$f' (2) = - 0.75680249 - 2 \sin (2)$

Step 12.3.2.1.3

Evaluate $\sin (2)$ .

$f' (2) = - 0.75680249 - 2 \cdot 0.90929742$

Step 12.3.2.1.4

Multiply $- 2$ by $0.90929742$ .

$f' (2) = - 0.75680249 - 1.81859485$

Step 12.3.2.2

Subtract $1.81859485$ from $- 0.75680249$ .

$f' (2) = - 2.57539734$

Step 12.3.2.3

The final answer is $- 2.57539734$ .

$- 2.57539734$

Step 12.4

Substitute any number, such as $5$ , from the interval $(π, 2 π)$ in the first derivative $\sin (2 x) - 2 \sin (x)$ to check if the result is negative or positive.

Tap for more steps...

Step 12.4.1

Replace the variable $x$ with $5$ in the expression.

$f' (5) = \sin (2 (5)) - 2 \sin (5)$

Step 12.4.2

Simplify the result.

Tap for more steps...

Step 12.4.2.1

Simplify each term.

Tap for more steps...

Step 12.4.2.1.1

Multiply $2$ by $5$ .

$f' (5) = \sin (10) - 2 \sin (5)$

Step 12.4.2.1.2

Evaluate $\sin (10)$ .

$f' (5) = - 0.54402111 - 2 \sin (5)$

Step 12.4.2.1.3

Evaluate $\sin (5)$ .

$f' (5) = - 0.54402111 - 2 \cdot - 0.95892427$

Step 12.4.2.1.4

Multiply $- 2$ by $- 0.95892427$ .

$f' (5) = - 0.54402111 + 1.91784854$

Step 12.4.2.2

Add $- 0.54402111$ and $1.91784854$ .

$f' (5) = 1.37382743$

Step 12.4.2.3

The final answer is $1.37382743$ .

$1.37382743$

Step 12.5

Substitute any number, such as $9$ , from the interval $(2 π, \infty)$ in the first derivative $\sin (2 x) - 2 \sin (x)$ to check if the result is negative or positive.

Tap for more steps...

Step 12.5.1

Replace the variable $x$ with $9$ in the expression.

$f' (9) = \sin (2 (9)) - 2 \sin (9)$

Step 12.5.2

Simplify the result.

Tap for more steps...

Step 12.5.2.1

Simplify each term.

Tap for more steps...

Step 12.5.2.1.1

Multiply $2$ by $9$ .

$f' (9) = \sin (18) - 2 \sin (9)$

Step 12.5.2.1.2

Evaluate $\sin (18)$ .

$f' (9) = - 0.75098724 - 2 \sin (9)$

Step 12.5.2.1.3

Evaluate $\sin (9)$ .

$f' (9) = - 0.75098724 - 2 \cdot 0.41211848$

Step 12.5.2.1.4

Multiply $- 2$ by $0.41211848$ .

$f' (9) = - 0.75098724 - 0.82423697$

Step 12.5.2.2

Subtract $0.82423697$ from $- 0.75098724$ .

$f' (9) = - 1.57522421$

Step 12.5.2.3

The final answer is $- 1.57522421$ .

$- 1.57522421$

Step 12.6

Since the first derivative changed signs from positive to negative around $x = 0$ , then $x = 0$ is a local maximum.

$x = 0$ is a local maximum

Step 12.7

Since the first derivative changed signs from negative to positive around $x = π$ , then $x = π$ is a local minimum.

$x = π$ is a local minimum

Step 12.8

Since the first derivative changed signs from positive to negative around $x = 2 π$ , then $x = 2 π$ is a local maximum.

$x = 2 π$ is a local maximum

Step 12.9

These are the local extrema for $f (x) = 2 \cos (x) + \sin^{2} (x)$ .

$x = 0$ is a local maximum

$x = π$ is a local minimum

$x = 2 π$ is a local maximum

$x = 0$ is a local maximum

$x = π$ is a local minimum

$x = 2 π$ is a local maximum

Step 13