Calculus Examples

$f (x) = 2 x^{3} - \frac{1}{x^{2}} + 4 \sqrt{x} + x - 11$

Step 1

By the Sum Rule, the derivative of $2 x^{3} - \frac{1}{x^{2}} + 4 \sqrt{x} + x - 11$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$ .

$\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 2

Evaluate $\frac{d}{d x} [2 x^{3}]$ .

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Step 2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

$2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 2.2

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

$2 (3 x^{2}) + \frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 2.3

Multiply $3$ by $2$ .

$6 x^{2} + \frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3

Evaluate $\frac{d}{d x} [- \frac{1}{x^{2}}]$ .

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Step 3.1

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 $f (x) = - 1$ and $g (x) = \frac{1}{x^{2}}$ .

$6 x^{2} - \frac{d}{d x} [\frac{1}{x^{2}}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.2

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$6 x^{2} - \frac{d}{d x} [{(x^{2})}^{- 1}] + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 3.3.1

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

$6 x^{2} - (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.3.2

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

$6 x^{2} - (- u^{- 2} \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.3.3

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

$6 x^{2} - (- {(x^{2})}^{- 2} \frac{d}{d x} [x^{2}]) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.4

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

$6 x^{2} - (- {(x^{2})}^{- 2} (2 x)) + \frac{1}{x^{2}} \frac{d}{d x} [- 1] + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.5

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

$6 x^{2} - (- {(x^{2})}^{- 2} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.6

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 3.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$6 x^{2} - (- x^{2 \cdot - 2} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.6.2

Multiply $2$ by $- 2$ .

$6 x^{2} - (- x^{- 4} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.7

Multiply $2$ by $- 1$ .

$6 x^{2} - (- 2 x^{- 4} x) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.8

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

$6 x^{2} - (- 2 (x^{1} x^{- 4})) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.9

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

$6 x^{2} - (- 2 x^{1 - 4}) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.10

Subtract $4$ from $1$ .

$6 x^{2} - (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.11

Multiply $- 2$ by $- 1$ .

$6 x^{2} + 2 x^{- 3} + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.12

Multiply $\frac{1}{x^{2}}$ by $0$ .

$6 x^{2} + 2 x^{- 3} + 0 + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 3.13

Add $2 x^{- 3}$ and $0$ .

$6 x^{2} + 2 x^{- 3} + \frac{d}{d x} [4 \sqrt{x}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4

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

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Step 4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$6 x^{2} + 2 x^{- 3} + \frac{d}{d x} [4 x^{\frac{1}{2}}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{\frac{1}{2}}$ with respect to $x$ is $4 \frac{d}{d x} [x^{\frac{1}{2}}]$ .

$6 x^{2} + 2 x^{- 3} + 4 \frac{d}{d x} [x^{\frac{1}{2}}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.3

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

$6 x^{2} + 2 x^{- 3} + 4 (\frac{1}{2} x^{\frac{1}{2} - 1}) + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$6 x^{2} + 2 x^{- 3} + 4 (\frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}}) + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.5

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

$6 x^{2} + 2 x^{- 3} + 4 (\frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}}) + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.6

Combine the numerators over the common denominator.

$6 x^{2} + 2 x^{- 3} + 4 (\frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}}) + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.7

Simplify the numerator.

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Step 4.7.1

Multiply $- 1$ by $2$ .

$6 x^{2} + 2 x^{- 3} + 4 (\frac{1}{2} x^{\frac{1 - 2}{2}}) + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.7.2

Subtract $2$ from $1$ .

$6 x^{2} + 2 x^{- 3} + 4 (\frac{1}{2} x^{\frac{- 1}{2}}) + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.8

Move the negative in front of the fraction.

$6 x^{2} + 2 x^{- 3} + 4 (\frac{1}{2} x^{- \frac{1}{2}}) + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.9

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

$6 x^{2} + 2 x^{- 3} + 4 \frac{x^{- \frac{1}{2}}}{2} + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.10

Combine $4$ and $\frac{x^{- \frac{1}{2}}}{2}$ .

$6 x^{2} + 2 x^{- 3} + \frac{4 x^{- \frac{1}{2}}}{2} + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.11

Move $x^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$6 x^{2} + 2 x^{- 3} + \frac{4}{2 x^{\frac{1}{2}}} + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.12

Factor $2$ out of $4$ .

$6 x^{2} + 2 x^{- 3} + \frac{2 \cdot 2}{2 x^{\frac{1}{2}}} + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.13

Cancel the common factors.

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Step 4.13.1

Factor $2$ out of $2 x^{\frac{1}{2}}$ .

$6 x^{2} + 2 x^{- 3} + \frac{2 \cdot 2}{2 (x^{\frac{1}{2}})} + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.13.2

Cancel the common factor.

$6 x^{2} + 2 x^{- 3} + \frac{2 \cdot 2}{2 x^{\frac{1}{2}}} + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 4.13.3

Rewrite the expression.

$6 x^{2} + 2 x^{- 3} + \frac{2}{x^{\frac{1}{2}}} + \frac{d}{d x} [x] + \frac{d}{d x} [- 11]$

Step 5

Differentiate.

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Step 5.1

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

$6 x^{2} + 2 x^{- 3} + \frac{2}{x^{\frac{1}{2}}} + 1 + \frac{d}{d x} [- 11]$

Step 5.2

Since $- 11$ is constant with respect to $x$ , the derivative of $- 11$ with respect to $x$ is $0$ .

$6 x^{2} + 2 x^{- 3} + \frac{2}{x^{\frac{1}{2}}} + 1 + 0$

Step 6

Simplify.

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Step 6.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$6 x^{2} + 2 \frac{1}{x^{3}} + \frac{2}{x^{\frac{1}{2}}} + 1 + 0$

Step 6.2

Combine terms.

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Step 6.2.1

Combine $2$ and $\frac{1}{x^{3}}$ .

$6 x^{2} + \frac{2}{x^{3}} + \frac{2}{x^{\frac{1}{2}}} + 1 + 0$

Step 6.2.2

Add $6 x^{2} + \frac{2}{x^{3}} + \frac{2}{x^{\frac{1}{2}}} + 1$ and $0$ .

$6 x^{2} + \frac{2}{x^{3}} + \frac{2}{x^{\frac{1}{2}}} + 1$