Find the derivative $$ y=\ln \sqrt{4 x-3} $$ Type answer $$ y^{\prime}= $$ (numerator)/(denominator)

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We start with
$$
y=\ln \sqrt{4x-3}.
$$
Recall that $$\sqrt{4x-3}=(4x-3)^{\frac{1}{2}}$$, so
$$
y=\frac{1}{2}\ln (4x-3).
$$
To differentiate $$y$$ with respect to $$x$$, we apply the chain rule. The derivative of $$\ln u$$ is $$\frac{1}{u} \frac{du}{dx}$$:

Let $$ u = 4x-3 $$, then
$$
\frac{d}{dx}\ln (4x-3) = \frac{1}{4x-3}\cdot 4.
$$
Thus,
$$
y^{\prime}=\frac{1}{2}\cdot \frac{4}{4x-3}=\frac{2}{4x-3}.
$$

Final answer:

$$
y^{\prime}=\frac{2}{4x-3}.
$$
$$ y^{\prime} = \frac{2}{4x - 3} $$

Find the derivative \( y=\ln \sqrt{4 x-3} \) Type answer \( y^{\prime}= \) (numerator)/(denominator)

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