Question

Find f′(x) using logarithmic differentiation , where f(x) =( e⁻³ˣ √(2x-5)) / (6 -5x)⁴

Answer 1

Given, $f(x)=\dfrac{e^{-3x}\sqrt{2x-5}}{(6-5x)^{4}}$. Applying logarithm on both sides,

$\ln f(x)=\ln\bigg\{\dfrac{e^{-3x}\sqrt{2x-5}}{(6-5x)^{4}}\bigg\}\\~ \\~~~~~~~~~~~~~=\ln (e^{-3x}\sqrt{2x-5}) - \ln (6-5x)^{4}~~~(\text{using}~\ln (\frac{a}{b})=\ln a- \ln b)\\~\\\ln f(x)=-3x+\dfrac{1}{2}\ln(2x-5)-4\ln(6-5x)\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\text{using $\ln (ab)=\ln a + \ln b, \ln e^{x} = x$ and $\ln a^{b} = b \ln a$) }\\~\\$

Differentiating with respect to x,

$\dfrac{f'(x)}{f(x)} = -3 + \dfrac{1}{2}\cdot \dfrac{1}{2x-5}\cdot 2 - 4 \cdot \dfrac{1}{6-5x}\cdot (-5)\\~\\ f'(x) = f(x)\bigg\{ -3 + \dfrac{1}{2x-5} + \dfrac{20}{6-5x}\bigg\}\\~\\ ~~~~~~~~~~=f(x)\bigg\{ \dfrac{-3(2x-5)(6-5x)+6-5x+20(2x-5)}{(2x-5)(6-5x)}\bigg\}\\~\\ f'(x)=\dfrac{e^{-3x}\sqrt{2x-5}}{(6-5x)^{4}}\bigg\{ \dfrac{30x^{2}-76x-4}{(2x-5)(6-5x)}\bigg\}\\~\\ f'(x)=2e^{-3x}\bigg\{ \dfrac{15x^{2}-38x-2}{\sqrt{2x-5}(6-5x)^{5}}\bigg\}$