Calculus Examples

$2 \sqrt{2 x - 2}$

Differentiate using the Constant Multiple Rule.

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

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

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

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

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

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To apply the Chain Rule, set $u$ as $2 x - 2$ .

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

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

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

Replace all occurrences of $u$ with $2 x - 2$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Simplify terms.

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Move the negative in front of the fraction.

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Combine fractions.

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Add $2$ and $0$ .

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

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

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

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