find the derivative.\n$\frac{d}{dx}leftlnleft(\frac{2x + 3}{2x - 3}\right)\right$\n$\frac{d}{dx}leftlnleft(\f…

Explanation:

Step1: Use the chain - rule

Let $u=\frac{2x + 3}{2x-3}$. Then we are finding $\frac{d}{dx}(\ln(u))$. By the chain - rule $\frac{d}{dx}(\ln(u))=\frac{1}{u}\cdot\frac{du}{dx}$.

Step2: Find $\frac{du}{dx}$ using the quotient - rule

The quotient - rule states that if $u=\frac{f(x)}{g(x)}$ where $f(x)=2x + 3$ and $g(x)=2x-3$, then $\frac{du}{dx}=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g(x)^2}$. Since $f^{\prime}(x)=2$ and $g^{\prime}(x)=2$, we have $\frac{du}{dx}=\frac{2(2x - 3)-2(2x + 3)}{(2x-3)^2}=\frac{4x-6-(4x + 6)}{(2x-3)^2}=\frac{4x-6 - 4x-6}{(2x-3)^2}=\frac{-12}{(2x-3)^2}$.

Step3: Substitute $u$ and $\frac{du}{dx}$ back into the chain - rule formula

Since $u = \frac{2x + 3}{2x-3}$ and $\frac{du}{dx}=\frac{-12}{(2x-3)^2}$, then $\frac{d}{dx}(\ln(u))=\frac{2x-3}{2x + 3}\cdot\frac{-12}{(2x-3)^2}=\frac{-12}{(2x + 3)(2x-3)}$.

Answer:

$\frac{-12}{(2x + 3)(2x-3)}$