Calculus Examples

$f (x) = 8 x^{2} + 8 \sin (2 x)$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $8 x^{2} + 8 \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [8 \sin (2 x)]$ .

$\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [8 \sin (2 x)]$

Step 1.2

Evaluate $\frac{d}{d x} [8 x^{2}]$ .

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Step 1.2.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{2}$ with respect to $x$ is $8 \frac{d}{d x} [x^{2}]$ .

$8 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [8 \sin (2 x)]$

Step 1.2.2

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

$8 (2 x) + \frac{d}{d x} [8 \sin (2 x)]$

Step 1.2.3

Multiply $2$ by $8$ .

$16 x + \frac{d}{d x} [8 \sin (2 x)]$

Step 1.3

Evaluate $\frac{d}{d x} [8 \sin (2 x)]$ .

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Step 1.3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 \sin (2 x)$ with respect to $x$ is $8 \frac{d}{d x} [\sin (2 x)]$ .

$16 x + 8 \frac{d}{d x} [\sin (2 x)]$

Step 1.3.2

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

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Step 1.3.2.1

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

$16 x + 8 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x])$

Step 1.3.2.2

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

$16 x + 8 (\cos (u) \frac{d}{d x} [2 x])$

Step 1.3.2.3

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

$16 x + 8 (\cos (2 x) \frac{d}{d x} [2 x])$

Step 1.3.3

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

$16 x + 8 (\cos (2 x) (2 \frac{d}{d x} [x]))$

Step 1.3.4

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

$16 x + 8 (\cos (2 x) (2 \cdot 1))$

Step 1.3.5

Multiply $2$ by $1$ .

$16 x + 8 (\cos (2 x) \cdot 2)$

Step 1.3.6

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

$16 x + 8 (2 \cdot \cos (2 x))$

Step 1.3.7

Multiply $2$ by $8$ .

$16 x + 16 \cos (2 x)$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $16 x + 16 \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [16 x] + \frac{d}{d x} [16 \cos (2 x)]$ .

$f'' (x) = \frac{d}{d x} (16 x) + \frac{d}{d x} (16 \cos (2 x))$

Step 2.2

Evaluate $\frac{d}{d x} [16 x]$ .

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Step 2.2.1

Since $16$ is constant with respect to $x$ , the derivative of $16 x$ with respect to $x$ is $16 \frac{d}{d x} [x]$ .

$f'' (x) = 16 \frac{d}{d x} (x) + \frac{d}{d x} (16 \cos (2 x))$

Step 2.2.2

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

$f'' (x) = 16 \cdot 1 + \frac{d}{d x} (16 \cos (2 x))$

Step 2.2.3

Multiply $16$ by $1$ .

$f'' (x) = 16 + \frac{d}{d x} (16 \cos (2 x))$

Step 2.3

Evaluate $\frac{d}{d x} [16 \cos (2 x)]$ .

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Step 2.3.1

Since $16$ is constant with respect to $x$ , the derivative of $16 \cos (2 x)$ with respect to $x$ is $16 \frac{d}{d x} [\cos (2 x)]$ .

$f'' (x) = 16 + 16 \frac{d}{d x} (\cos (2 x))$

Step 2.3.2

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 $g (x) = 2 x$ .

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Step 2.3.2.1

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

$f'' (x) = 16 + 16 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (2 x))$

Step 2.3.2.2

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

$f'' (x) = 16 + 16 (- \sin (u) \frac{d}{d x} (2 x))$

Step 2.3.2.3

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

$f'' (x) = 16 + 16 (- \sin (2 x) \frac{d}{d x} (2 x))$

Step 2.3.3

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

$f'' (x) = 16 + 16 (- \sin (2 x) (2 \frac{d}{d x} (x)))$

Step 2.3.4

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

$f'' (x) = 16 + 16 (- \sin (2 x) (2 \cdot 1))$

Step 2.3.5

Multiply $2$ by $1$ .

$f'' (x) = 16 + 16 (- \sin (2 x) \cdot 2)$

Step 2.3.6

Multiply $2$ by $- 1$ .

$f'' (x) = 16 + 16 (- 2 \sin (2 x))$

Step 2.3.7

Multiply $- 2$ by $16$ .

$f'' (x) = 16 - 32 \sin (2 x)$

Step 3

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

$16 x + 16 \cos (2 x) = 0$

Step 4

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx - 0.51493326$

Step 5

Evaluate the second derivative at $x = - 0.51493326$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$16 - 32 \sin (2 (- 0.51493326))$

Step 6

Multiply $2$ by $- 0.51493326$ .

$16 - 32 \sin (- 1.02986652)$

Step 7

$x = - 0.51493326$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 0.51493326$ is a local minimum

Step 8

Find the y-value when $x = - 0.51493326$ .

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Step 8.1

Replace the variable $x$ with $- 0.51493326$ in the expression.

$f (- 0.51493326) = 8 {(- 0.51493326)}^{2} + 8 \sin (2 (- 0.51493326))$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $- 0.51493326$ to the power of $2$ .

$f (- 0.51493326) = 8 \cdot 0.26515626 + 8 \sin (2 (- 0.51493326))$

Step 8.2.1.2

Multiply $8$ by $0.26515626$ .

$f (- 0.51493326) = 2.12125013 + 8 \sin (2 (- 0.51493326))$

Step 8.2.1.3

Multiply $2$ by $- 0.51493326$ .

$f (- 0.51493326) = 2.12125013 + 8 \sin (- 1.02986652)$

Step 8.2.2

Add $2.12125013$ and $8 \sin (- 1.02986652)$ .

$f (- 0.51493326) = - 4.73659201$

Step 8.2.3

The final answer is $- 4.73659201$ .

$y = - 4.73659201$

Step 9

These are the local extrema for $f (x) = 8 x^{2} + 8 \sin (2 x)$ .

$(- 0.51493326, - 4.73659201)$ is a local minima

Step 10