If f(x)=log _eleft(e^{2 x}left(frac{3 x+5}{5- | vedclass

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(D) Given, $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}\right)$ Using the property $\log(ab) = \log a + \log b$, we get: $

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$\frac{13}{4}$

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(D) Given, $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}\right)$ Using the property $\log(ab) = \log a + \log b$, we get: $f(x)=\log _e e^{2 x}+\log _e\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}$ $f(x)=2 x+\frac{2}{3} \log _e\left(\frac{3 x+5}{5-3 x}\right)$ Using $\log(\frac{a}{b}) = \log a - \log b$: $f(x)=2 x+\frac{2}{3}\left(\log _e(3 x+5)-\log _e(5-3 x)\right)$ Differentiating with respect to $x$: $\frac{d f}{d x}=2+\frac{2}{3}\left(\frac{3}{3 x+5}-\frac{-3}{5-3 x}\right)$ $\frac{d f}{d x}=2+2\left(\frac{1}{3 x+5}+\frac{1}{5-3 x}\right)$ At $x=1$: $\left(\frac{d f}{d x}\right)_{x=1}=2+2\left(\frac{1}{3(1)+5}+\frac{1}{5-3(1)}\right)$ $=2+2\left(\frac{1}{8}+\frac{1}{2}\right)=2+2\left(\frac{1+4}{8}\right)=2+2\left(\frac{5}{8}\right)=2+\frac{5}{4}=\frac{13}{4}$

If $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}\right)$, $x \neq \frac{-5}{3}, \frac{5}{3}$, then the value of $\frac{d f}{d x}$ at $x=1$ is

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