Find dy/dx y = square root of natural log of 4x

$y = \sqrt{\ln (4 x)}$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{\ln (4 x)}$ as ${\ln (4 x)}^{\frac{1}{2}}$ .

$y = {(\ln (4 x))}^{\frac{1}{2}}$

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ({(\ln (4 x))}^{\frac{1}{2}})$

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Differentiate the right side of the equation.

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 $g (x) = \ln (4 x)$ .

To apply the Chain Rule, set $u_{1}$ as $\ln (4 x)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{\frac{1}{2}}] \frac{d}{d x} [\ln (4 x)]$

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

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

Replace all occurrences of $u_{1}$ with $\ln (4 x)$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Combine fractions.

Move the negative in front of the fraction.

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

Combine $\frac{1}{2}$ and ${\ln (4 x)}^{- \frac{1}{2}}$ .

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

Move ${\ln (4 x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 4 x$ .

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$\frac{1}{2 {\ln (4 x)}^{\frac{1}{2}}} (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [4 x])$

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\frac{1}{2 {\ln (4 x)}^{\frac{1}{2}}} (\frac{1}{u_{2}} \frac{d}{d x} [4 x])$

Replace all occurrences of $u_{2}$ with $4 x$ .

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

Differentiate using the Constant Multiple Rule.

Multiply $\frac{1}{4 x}$ by $\frac{1}{2 {\ln (4 x)}^{\frac{1}{2}}}$ .

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

Multiply $2$ by $4$ .

$\frac{1}{8 x {\ln (4 x)}^{\frac{1}{2}}} \frac{d}{d x} [4 x]$

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

$\frac{1}{8 x {\ln (4 x)}^{\frac{1}{2}}} (4 \frac{d}{d x} [x])$

Simplify terms.

Combine $4$ and $\frac{1}{8 x {\ln (4 x)}^{\frac{1}{2}}}$ .

$\frac{4}{8 x {\ln (4 x)}^{\frac{1}{2}}} \frac{d}{d x} [x]$

Factor $4$ out of $4$ .

$\frac{4 (1)}{8 x {\ln (4 x)}^{\frac{1}{2}}} \frac{d}{d x} [x]$

Cancel the common factors.

Factor $4$ out of $8 x {\ln (4 x)}^{\frac{1}{2}}$ .

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

Cancel the common factor.

$\frac{4 \cdot 1}{4 (2 x {\ln (4 x)}^{\frac{1}{2}})} \frac{d}{d x} [x]$

Rewrite the expression.

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

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

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

Multiply $\frac{1}{2 x {\ln (4 x)}^{\frac{1}{2}}}$ by $1$ .

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

Reform the equation by setting the left side equal to the right side.

$y' = \frac{1}{2 x {(\ln (4 x))}^{\frac{1}{2}}}$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{1}{2 x {(\ln (4 x))}^{\frac{1}{2}}}$

Find dy/dx y = square root of natural log of 4x

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