Calculus Examples

$y = \frac{1}{x^{2}} - \frac{1}{x^{3}}$

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $x^{2}$ .

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u_{1}$ with $x^{2}$ .

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

Step 2.3

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

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

Multiply $2$ by $- 2$ .

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

Step 2.5

Multiply $2$ by $- 1$ .

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

Subtract $4$ from $1$ .

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

Step 3

Evaluate $\frac{d}{d x} [- \frac{1}{x^{3}}]$ .

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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 $g (x) = \frac{1}{x^{3}}$ .

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

Step 3.2

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

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

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

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Step 3.3.1

To apply the Chain Rule, set $u_{2}$ as $x^{3}$ .

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

Step 3.3.2

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

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

Step 3.3.3

Replace all occurrences of $u_{2}$ with $x^{3}$ .

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

Step 3.4

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

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

Step 3.5

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

$- 2 x^{- 3} - (- {(x^{3})}^{- 2} (3 x^{2})) + \frac{1}{x^{3}} \cdot 0$

Step 3.6

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

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Step 3.6.1

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

$- 2 x^{- 3} - (- x^{3 \cdot - 2} (3 x^{2})) + \frac{1}{x^{3}} \cdot 0$

Step 3.6.2

Multiply $3$ by $- 2$ .

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

Step 3.7

Multiply $3$ by $- 1$ .

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

Step 3.8

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

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Step 3.8.1

Move $x^{2}$ .

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

Step 3.8.2

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

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

Step 3.8.3

Subtract $6$ from $2$ .

$- 2 x^{- 3} - (- 3 x^{- 4}) + \frac{1}{x^{3}} \cdot 0$

Step 3.9

Multiply $- 3$ by $- 1$ .

$- 2 x^{- 3} + 3 x^{- 4} + \frac{1}{x^{3}} \cdot 0$

Step 3.10

Multiply $\frac{1}{x^{3}}$ by $0$ .

$- 2 x^{- 3} + 3 x^{- 4} + 0$

Step 3.11

Add $3 x^{- 4}$ and $0$ .

$- 2 x^{- 3} + 3 x^{- 4}$

Step 4

Simplify.

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Step 4.1

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

$- 2 \frac{1}{x^{3}} + 3 x^{- 4}$

Step 4.2

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

$- 2 \frac{1}{x^{3}} + 3 \frac{1}{x^{4}}$

Step 4.3

Combine terms.

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Step 4.3.1

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

$\frac{- 2}{x^{3}} + 3 \frac{1}{x^{4}}$

Step 4.3.2

Move the negative in front of the fraction.

$- \frac{2}{x^{3}} + 3 \frac{1}{x^{4}}$

Step 4.3.3

Combine $3$ and $\frac{1}{x^{4}}$ .

$- \frac{2}{x^{3}} + \frac{3}{x^{4}}$