Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Multiply $- 1$ by $2$ .

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

Step 3

Evaluate $\frac{d}{d x} [\frac{1}{4 x^{2}}]$ .

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Step 3.1

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

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

Step 3.2

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

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

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 $g (x) = x^{2}$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

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

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

Step 3.5

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

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Step 3.5.1

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

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

Step 3.5.2

Multiply $2$ by $- 2$ .

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

Step 3.6

Multiply $2$ by $- 1$ .

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

Subtract $4$ from $1$ .

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

Step 3.10

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

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

Step 3.11

Combine $\frac{- 2}{4}$ and $x^{- 3}$ .

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

Step 3.12

Move $x^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.13

Cancel the common factor of $- 2$ and $4$ .

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Step 3.13.1

Factor $2$ out of $- 2$ .

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

Step 3.13.2

Cancel the common factors.

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Step 3.13.2.1

Factor $2$ out of $4 x^{3}$ .

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

Step 3.13.2.2

Cancel the common factor.

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

Step 3.13.2.3

Rewrite the expression.

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

Step 3.14

Move the negative in front of the fraction.

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

Step 4

Simplify.

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Step 4.1

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

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

Step 4.2

Combine terms.

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Step 4.2.1

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

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

Step 4.2.2

Move the negative in front of the fraction.

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