Let $$ f(x)=\ln \left(\sqrt{\frac{8 x-9}{3 x+2}}\right) $$ $$ f^{\prime}(x)=\square $$ Question Help: VVideo Submit Question

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We start with
$$
f(x)=\ln \left(\sqrt{\frac{8x-9}{3x+2}}\right).
$$
Since the logarithm of a square root can be rewritten using the property $$\ln(a^b)=b\ln(a)$$, we have
$$
f(x)=\frac{1}{2}\ln \left(\frac{8x-9}{3x+2}\right).
$$

Next, use the logarithm property $$\ln\left(\frac{A}{B}\right)=\ln A-\ln B$$ to obtain
$$
f(x)=\frac{1}{2}\Bigl(\ln(8x-9)-\ln(3x+2)\Bigr).
$$

We now differentiate $$f(x)$$ term-by-term. Recall that the derivative of $$\ln(u(x))$$ with respect to $$x$$ is
$$
\frac{u'(x)}{u(x)}.
$$

1. Differentiate $$\ln(8x-9)$$:
   - Let $$u(x)=8x-9$$. Then $$u'(x)=8$$. Thus,
     $$
     \frac{d}{dx}\ln(8x-9)=\frac{8}{8x-9}.
     $$

2. Differentiate $$\ln(3x+2)$$:
   - Let $$v(x)=3x+2$$. Then $$v'(x)=3$$. Thus,
     $$
     \frac{d}{dx}\ln(3x+2)=\frac{3}{3x+2}.
     $$

Now, putting these results together, we have
$$
f'(x)=\frac{1}{2}\left(\frac{8}{8x-9}-\frac{3}{3x+2}\right).
$$

We can also combine the two fractions:
$$
f'(x)=\frac{1}{2}\left(\frac{8(3x+2)-3(8x-9)}{(8x-9)(3x+2)}\right).
$$

Simplify the numerator:
$$
8(3x+2)=24x+16 \quad \text{and} \quad 3(8x-9)=24x-27.
$$
Thus,
$$
8(3x+2)-3(8x-9)=(24x+16)-(24x-27)=24x+16-24x+27=43.
$$

So the derivative becomes:
$$
f'(x)=\frac{1}{2}\cdot\frac{43}{(8x-9)(3x+2)}=\frac{43}{2(8x-9)(3x+2)}.
$$

Thus,
$$
\boxed{f'(x)=\frac{43}{2(8x-9)(3x+2)}}.
$$
$$ f'(x) = \frac{43}{2(8x - 9)(3x + 2)} $$

Let \( f(x)=\ln \left(\sqrt{\frac{8 x-9}{3 x+2}}\right) \) \( f^{\prime}(x)=\square \) Question Help: VVideo Submit Question

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