Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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 $g (x) = \sin (x)$ .

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Reorder the factors of $2 \sin (x) \cos (x)$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

Apply the sine double-angle identity.

$\sin (2 x)$

Step 2

Find the second derivative of the function.

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Step 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 $g (x) = 2 x$ .

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Simplify the expression.

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Step 2.2.3.1

Multiply $2$ by $1$ .

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

Step 2.2.3.2

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

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

Step 3

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

$\sin (2 x) = 0$

Step 4

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

$2 x = \arcsin (0)$

Step 5

Simplify the right side.

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Step 5.1

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

$2 x = 0$

Step 6

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 6.1

Divide each term in $2 x = 0$ by $2$ .

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

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $2$ .

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Step 6.2.1.1

Cancel the common factor.

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

Step 6.2.1.2

Divide $x$ by $1$ .

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

Step 6.3

Simplify the right side.

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Step 6.3.1

Divide $0$ by $2$ .

$x = 0$

Step 7

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.

$2 x = π - 0$

Step 8

Solve for $x$ .

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Step 8.1

Simplify.

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Step 8.1.1

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Step 8.1.2

Add $π$ and $0$ .

$2 x = π$

Step 8.2

Divide each term in $2 x = π$ by $2$ and simplify.

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Step 8.2.1

Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 8.2.2

Simplify the left side.

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Step 8.2.2.1

Cancel the common factor of $2$ .

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Step 8.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 8.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 9

The solution to the equation $2 x = 0$ .

$x = 0, \frac{π}{2}$

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.

$2 \cos (2 (0))$

Step 11

Evaluate the second derivative.

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Step 11.1

Multiply $2$ by $0$ .

$2 \cos (0)$

Step 11.2

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

$2 \cdot 1$

Step 11.3

Multiply $2$ by $1$ .

$2$

Step 12

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 13

Find the y-value when $x = 0$ .

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Step 13.1

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

$f (0) = \sin^{2} (0)$

Step 13.2

Simplify the result.

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Step 13.2.1

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

$f (0) = 0^{2}$

Step 13.2.2

Raising $0$ to any positive power yields $0$ .

$f (0) = 0$

Step 13.2.3

The final answer is $0$ .

$y = 0$

Step 14

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.

$2 \cos (2 (\frac{π}{2}))$

Step 15

Evaluate the second derivative.

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Step 15.1

Cancel the common factor of $2$ .

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Step 15.1.1

Cancel the common factor.

$2 \cos (2 \frac{π}{2})$

Step 15.1.2

Rewrite the expression.

$2 \cos (π)$

Step 15.2

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

$2 (- \cos (0))$

Step 15.3

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

$2 (- 1 \cdot 1)$

Step 15.4

Multiply $2 (- 1 \cdot 1)$ .

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Step 15.4.1

Multiply $- 1$ by $1$ .

$2 \cdot - 1$

Step 15.4.2

Multiply $2$ by $- 1$ .

$- 2$

Step 16

$x = \frac{π}{2}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{2}$ is a local maximum

Step 17

Find the y-value when $x = \frac{π}{2}$ .

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Step 17.1

Replace the variable $x$ with $\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = \sin^{2} (\frac{π}{2})$

Step 17.2

Simplify the result.

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Step 17.2.1

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

$f (\frac{π}{2}) = 1^{2}$

Step 17.2.2

One to any power is one.

$f (\frac{π}{2}) = 1$

Step 17.2.3

The final answer is $1$ .

$y = 1$

Step 18

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

$(0, 0)$ is a local minima

$(\frac{π}{2}, 1)$ is a local maxima

Step 19