Calculus Examples

$y = (6 x^{2} - 1) (3 x^{3} + x)$

Step 1

Find the first derivative.

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 6 x^{2} - 1$ and $g (x) = 3 x^{3} + x$ .

$f' (x) = (6 x^{2} - 1) \frac{d}{d x} (3 x^{3} + x) + (3 x^{3} + x) \frac{d}{d x} (6 x^{2} - 1)$

Differentiate.

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By the Sum Rule, the derivative of $3 x^{3} + x$ with respect to $x$ is $\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [x]$ .

$f' (x) = (6 x^{2} - 1) (\frac{d}{d x} (3 x^{3}) + \frac{d}{d x} (x)) + (3 x^{3} + x) \frac{d}{d x} (6 x^{2} - 1)$

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{3}$ with respect to $x$ is $3 \frac{d}{d x} [x^{3}]$ .

$f' (x) = (6 x^{2} - 1) (3 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (x)) + (3 x^{3} + x) \frac{d}{d x} (6 x^{2} - 1)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = (6 x^{2} - 1) (3 (3 x^{2}) + \frac{d}{d x} (x)) + (3 x^{3} + x) \frac{d}{d x} (6 x^{2} - 1)$

Multiply $3$ by $3$ .

$f' (x) = (6 x^{2} - 1) (9 x^{2} + \frac{d}{d x} (x)) + (3 x^{3} + x) \frac{d}{d x} (6 x^{2} - 1)$

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

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (3 x^{3} + x) \frac{d}{d x} (6 x^{2} - 1)$

By the Sum Rule, the derivative of $6 x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 1]$ .

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (3 x^{3} + x) (\frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (- 1))$

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (3 x^{3} + x) (6 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 1))$

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

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (3 x^{3} + x) (6 (2 x) + \frac{d}{d x} (- 1))$

Multiply $2$ by $6$ .

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (3 x^{3} + x) (12 x + \frac{d}{d x} (- 1))$

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (3 x^{3} + x) (12 x + 0)$

Simplify the expression.

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Add $12 x$ and $0$ .

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (3 x^{3} + x) (12 x)$

Move $12$ to the left of $3 x^{3} + x$ .

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + 12 \cdot ((3 x^{3} + x) x)$

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + 12 \cdot ((3 x^{3} + x) x)$

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + 12 \cdot ((3 x^{3} + x) x)$

Simplify.

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Apply the distributive property.

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + (12 (3 x^{3}) + 12 x) x$

Apply the distributive property.

$f' (x) = (6 x^{2} - 1) (9 x^{2} + 1) + 12 (3 x^{3}) x + 12 x \cdot x$

Apply the distributive property.

$f' (x) = (6 x^{2} - 1) (9 x^{2}) + (6 x^{2} - 1) \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Apply the distributive property.

$f' (x) = 6 x^{2} (9 x^{2}) - 1 (9 x^{2}) + (6 x^{2} - 1) \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Apply the distributive property.

$f' (x) = 6 x^{2} (9 x^{2}) - 1 (9 x^{2}) + 6 x^{2} \cdot 1 - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Combine terms.

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Multiply $9$ by $6$ .

$f' (x) = 54 x^{2} x^{2} - 1 (9 x^{2}) + 6 x^{2} \cdot 1 - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Move $x^{2}$ .

$f' (x) = 54 (x^{2} x^{2}) - 1 (9 x^{2}) + 6 x^{2} \cdot 1 - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (x) = 54 x^{2 + 2} - 1 (9 x^{2}) + 6 x^{2} \cdot 1 - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Add $2$ and $2$ .

$f' (x) = 54 x^{4} - 1 (9 x^{2}) + 6 x^{2} \cdot 1 - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

$f' (x) = 54 x^{4} - 1 (9 x^{2}) + 6 x^{2} \cdot 1 - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Multiply $9$ by $- 1$ .

$f' (x) = 54 x^{4} - 9 x^{2} + 6 x^{2} \cdot 1 - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Multiply $6$ by $1$ .

$f' (x) = 54 x^{4} - 9 x^{2} + 6 x^{2} - 1 \cdot 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Multiply $- 1$ by $1$ .

$f' (x) = 54 x^{4} - 9 x^{2} + 6 x^{2} - 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Add $- 9 x^{2}$ and $6 x^{2}$ .

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 12 (3 x^{3}) x + 12 x \cdot x$

Multiply $3$ by $12$ .

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 x^{3} x + 12 x \cdot x$

Raise $x$ to the power of $1$ .

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 (x \cdot x^{3}) + 12 x \cdot x$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 x^{1 + 3} + 12 x \cdot x$

Add $1$ and $3$ .

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 x^{4} + 12 x \cdot x$

Raise $x$ to the power of $1$ .

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 x^{4} + 12 (x \cdot x)$

Raise $x$ to the power of $1$ .

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 x^{4} + 12 (x \cdot x)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 x^{4} + 12 x^{1 + 1}$

Add $1$ and $1$ .

$f' (x) = 54 x^{4} - 3 x^{2} - 1 + 36 x^{4} + 12 x^{2}$

Add $54 x^{4}$ and $36 x^{4}$ .

$f' (x) = 90 x^{4} - 3 x^{2} - 1 + 12 x^{2}$

Add $- 3 x^{2}$ and $12 x^{2}$ .

$f' (x) = 90 x^{4} + 9 x^{2} - 1$

$f' (x) = 90 x^{4} + 9 x^{2} - 1$

$f' (x) = 90 x^{4} + 9 x^{2} - 1$

$f' (x) = 90 x^{4} + 9 x^{2} - 1$

Step 2

Find the second derivative.

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By the Sum Rule, the derivative of $90 x^{4} + 9 x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [90 x^{4}] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- 1]$ .

$f'' (x) = \frac{d}{d x} (90 x^{4}) + \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (- 1)$

Evaluate $\frac{d}{d x} [90 x^{4}]$ .

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Since $90$ is constant with respect to $x$ , the derivative of $90 x^{4}$ with respect to $x$ is $90 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 90 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (- 1)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 90 (4 x^{3}) + \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (- 1)$

Multiply $4$ by $90$ .

$f'' (x) = 360 x^{3} + \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (- 1)$

$f'' (x) = 360 x^{3} + \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (- 1)$

Evaluate $\frac{d}{d x} [9 x^{2}]$ .

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Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 360 x^{3} + 9 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 1)$

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

$f'' (x) = 360 x^{3} + 9 (2 x) + \frac{d}{d x} (- 1)$

Multiply $2$ by $9$ .

$f'' (x) = 360 x^{3} + 18 x + \frac{d}{d x} (- 1)$

$f'' (x) = 360 x^{3} + 18 x + \frac{d}{d x} (- 1)$

Differentiate using the Constant Rule.

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Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = 360 x^{3} + 18 x + 0$

Add $360 x^{3} + 18 x$ and $0$ .

$f'' (x) = 360 x^{3} + 18 x$

$f'' (x) = 360 x^{3} + 18 x$

$f'' (x) = 360 x^{3} + 18 x$

Step 3

Find the third derivative.

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By the Sum Rule, the derivative of $360 x^{3} + 18 x$ with respect to $x$ is $\frac{d}{d x} [360 x^{3}] + \frac{d}{d x} [18 x]$ .

$f''' (x) = \frac{d}{d x} (360 x^{3}) + \frac{d}{d x} (18 x)$

Evaluate $\frac{d}{d x} [360 x^{3}]$ .

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Since $360$ is constant with respect to $x$ , the derivative of $360 x^{3}$ with respect to $x$ is $360 \frac{d}{d x} [x^{3}]$ .

$f''' (x) = 360 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (18 x)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f''' (x) = 360 (3 x^{2}) + \frac{d}{d x} (18 x)$

Multiply $3$ by $360$ .

$f''' (x) = 1080 x^{2} + \frac{d}{d x} (18 x)$

$f''' (x) = 1080 x^{2} + \frac{d}{d x} (18 x)$

Evaluate $\frac{d}{d x} [18 x]$ .

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Since $18$ is constant with respect to $x$ , the derivative of $18 x$ with respect to $x$ is $18 \frac{d}{d x} [x]$ .

$f''' (x) = 1080 x^{2} + 18 \frac{d}{d x} (x)$

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

$f''' (x) = 1080 x^{2} + 18 \cdot 1$

Multiply $18$ by $1$ .

$f''' (x) = 1080 x^{2} + 18$

$f''' (x) = 1080 x^{2} + 18$

$f''' (x) = 1080 x^{2} + 18$

Step 4

Find the fourth derivative.

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By the Sum Rule, the derivative of $1080 x^{2} + 18$ with respect to $x$ is $\frac{d}{d x} [1080 x^{2}] + \frac{d}{d x} [18]$ .

$f^{4} (x) = \frac{d}{d x} (1080 x^{2}) + \frac{d}{d x} (18)$

Evaluate $\frac{d}{d x} [1080 x^{2}]$ .

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Since $1080$ is constant with respect to $x$ , the derivative of $1080 x^{2}$ with respect to $x$ is $1080 \frac{d}{d x} [x^{2}]$ .

$f^{4} (x) = 1080 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (18)$

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

$f^{4} (x) = 1080 (2 x) + \frac{d}{d x} (18)$

Multiply $2$ by $1080$ .

$f^{4} (x) = 2160 x + \frac{d}{d x} (18)$

$f^{4} (x) = 2160 x + \frac{d}{d x} (18)$

Differentiate using the Constant Rule.

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Since $18$ is constant with respect to $x$ , the derivative of $18$ with respect to $x$ is $0$ .

$f^{4} (x) = 2160 x + 0$

Add $2160 x$ and $0$ .

$f^{4} (x) = 2160 x$

$f^{4} (x) = 2160 x$

$f^{4} (x) = 2160 x$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$