Calculus Examples

$x + \frac{32}{x^{2}}$

Step 1

Find the first derivative.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $x + \frac{32}{x^{2}}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{32}{x^{2}}]$ .

$f' (x) = \frac{d}{d x} (x) + \frac{d}{d x} (\frac{32}{x^{2}})$

Step 1.1.2

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

$f' (x) = 1 + \frac{d}{d x} (\frac{32}{x^{2}})$

Step 1.2

Evaluate $\frac{d}{d x} [\frac{32}{x^{2}}]$ .

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Step 1.2.1

Since $32$ is constant with respect to $x$ , the derivative of $\frac{32}{x^{2}}$ with respect to $x$ is $32 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f' (x) = 1 + 32 \frac{d}{d x} (\frac{1}{x^{2}})$

Step 1.2.2

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

$f' (x) = 1 + 32 \frac{d}{d x} ({(x^{2})}^{- 1})$

Step 1.2.3

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 1.2.3.1

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

$f' (x) = 1 + 32 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2}))$

Step 1.2.3.2

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

$f' (x) = 1 + 32 (- u^{- 2} \frac{d}{d x} x^{2})$

Step 1.2.3.3

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

$f' (x) = 1 + 32 (- {(x^{2})}^{- 2} \frac{d}{d x} x^{2})$

Step 1.2.4

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

$f' (x) = 1 + 32 (- {(x^{2})}^{- 2} (2 x))$

Step 1.2.5

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

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Step 1.2.5.1

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

$f' (x) = 1 + 32 (- x^{2 \cdot - 2} (2 x))$

Step 1.2.5.2

Multiply $2$ by $- 2$ .

$f' (x) = 1 + 32 (- x^{- 4} (2 x))$

Step 1.2.6

Multiply $2$ by $- 1$ .

$f' (x) = 1 + 32 (- 2 x^{- 4} x)$

Step 1.2.7

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

$f' (x) = 1 + 32 (- 2 (x \cdot x^{- 4}))$

Step 1.2.8

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

$f' (x) = 1 + 32 (- 2 x^{1 - 4})$

Step 1.2.9

Subtract $4$ from $1$ .

$f' (x) = 1 + 32 (- 2 x^{- 3})$

Step 1.2.10

Multiply $- 2$ by $32$ .

$f' (x) = 1 - 64 x^{- 3}$

Step 1.3

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

$f' (x) = 1 - 64 \frac{1}{x^{3}}$

Step 1.4

Simplify.

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Step 1.4.1

Combine terms.

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Step 1.4.1.1

Combine $- 64$ and $\frac{1}{x^{3}}$ .

$f' (x) = 1 + \frac{- 64}{x^{3}}$

Step 1.4.1.2

Move the negative in front of the fraction.

$f' (x) = 1 - \frac{64}{x^{3}}$

Step 1.4.2

Reorder terms.

$f' (x) = - \frac{64}{x^{3}} + 1$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $- \frac{64}{x^{3}} + 1$ with respect to $x$ is $\frac{d}{d x} [- \frac{64}{x^{3}}] + \frac{d}{d x} [1]$ .

$f'' (x) = \frac{d}{d x} (- \frac{64}{x^{3}}) + \frac{d}{d x} (1)$

Step 2.2

Evaluate $\frac{d}{d x} [- \frac{64}{x^{3}}]$ .

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Step 2.2.1

Since $- 64$ is constant with respect to $x$ , the derivative of $- \frac{64}{x^{3}}$ with respect to $x$ is $- 64 \frac{d}{d x} [\frac{1}{x^{3}}]$ .

$f'' (x) = - 64 \frac{d}{d x} \frac{1}{x^{3}} + \frac{d}{d x} (1)$

Step 2.2.2

Rewrite $\frac{1}{x^{3}}$ as ${(x^{3})}^{- 1}$ .

$f'' (x) = - 64 \frac{d}{d x} {(x^{3})}^{- 1} + \frac{d}{d x} (1)$

Step 2.2.3

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 $g (x) = x^{3}$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u$ as $x^{3}$ .

$f'' (x) = - 64 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{3})) + \frac{d}{d x} (1)$

Step 2.2.3.2

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

$f'' (x) = - 64 (- u^{- 2} \frac{d}{d x} x^{3}) + \frac{d}{d x} (1)$

Step 2.2.3.3

Replace all occurrences of $u$ with $x^{3}$ .

$f'' (x) = - 64 (- {(x^{3})}^{- 2} \frac{d}{d x} x^{3}) + \frac{d}{d x} (1)$

Step 2.2.4

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

$f'' (x) = - 64 (- {(x^{3})}^{- 2} (3 x^{2})) + \frac{d}{d x} (1)$

Step 2.2.5

Multiply the exponents in ${(x^{3})}^{- 2}$ .

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Step 2.2.5.1

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

$f'' (x) = - 64 (- x^{3 \cdot - 2} (3 x^{2})) + \frac{d}{d x} (1)$

Step 2.2.5.2

Multiply $3$ by $- 2$ .

$f'' (x) = - 64 (- x^{- 6} (3 x^{2})) + \frac{d}{d x} (1)$

Step 2.2.6

Multiply $3$ by $- 1$ .

$f'' (x) = - 64 (- 3 x^{- 6} x^{2}) + \frac{d}{d x} (1)$

Step 2.2.7

Multiply $x^{- 6}$ by $x^{2}$ by adding the exponents.

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Step 2.2.7.1

Move $x^{2}$ .

$f'' (x) = - 64 (- 3 (x^{2} x^{- 6})) + \frac{d}{d x} (1)$

Step 2.2.7.2

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

$f'' (x) = - 64 (- 3 x^{2 - 6}) + \frac{d}{d x} (1)$

Step 2.2.7.3

Subtract $6$ from $2$ .

$f'' (x) = - 64 (- 3 x^{- 4}) + \frac{d}{d x} (1)$

Step 2.2.8

Multiply $- 3$ by $- 64$ .

$f'' (x) = 192 x^{- 4} + \frac{d}{d x} (1)$

Step 2.3

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

$f'' (x) = 192 x^{- 4} + 0$

Step 2.4

Simplify.

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Step 2.4.1

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

$f'' (x) = 192 (\frac{1}{x^{4}}) + 0$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $192$ and $\frac{1}{x^{4}}$ .

$f'' (x) = \frac{192}{x^{4}} + 0$

Step 2.4.2.2

Add $\frac{192}{x^{4}}$ and $0$ .

$f'' (x) = \frac{192}{x^{4}}$