Calculus Examples

$\frac{1}{\sqrt{g (2 x)}}$

Step 1

Apply basic rules of exponents.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Multiply the exponents in ${({g (2 x)}^{\frac{1}{2}})}^{- 1}$ .

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Move the negative in front of the fraction.

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

Step 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 $g (x) = g (2 x)$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $g (2 x)$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $g (2 x)$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $- 1$ .

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

Step 7

Move the negative in front of the fraction.

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

Step 8

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

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

Step 9

Move ${g (2 x)}^{- \frac{3}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 10

Combine $- \frac{1}{2 {g (2 x)}^{\frac{3}{2}}}$ and $\frac{d}{d x} [g (2 x)]$ .

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