Can some one help me with this problem: using logarithmic differentiation find the derivative of $$\frac {e^{2x} (x^3-2)^4}{x(3e^{5x} +1)}.$$ I got stumped its a fraction, has e in it, and I'm supposed to use logarithmic differentiation.
2026-05-14 07:37:24.1778744244
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logarithmic differentiation of fraction
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Don't worry: logarithmic differentiation of a fraction $$ F(x)=\frac{f(x)}{g(x)} $$ consists in writing, formally, $\log F(x)=\log f(x)-\log g(x)$ and so $$ \frac{F'(x)}{F(x)}=\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)} $$ If you have more terms, it's no different. In your case $f(x)=f_1(x)f_2(x)$ and $g(x)=g_1(x)g_2(x)$ (write down the definitions for these functions), so $\log F(x)=\log f_1(x)+\log f_2(x)-\log g_1(x)-\log g_2(x)$ and $$ \frac{F'(x)}{F(x)}=\frac{f_1'(x)}{f_1(x)}+\frac{f_2'(x)}{f_2(x)} -\frac{g_1'(x)}{g_1(x)}-\frac{g_2'(x)}{g_2(x)} $$ Now just compute the derivatives and simplify.
Take natural logarithm from both sides of the given function: $$y=\frac {e^{2x} (x^3-2)^4}{x(3e^{5x} +1)} \Rightarrow \\ \begin{align}\ln y&=\ln \frac {e^{2x} (x^3-2)^4}{x(3e^{5x} +1)}=\\ &=\ln (e^{2x} (x^3-2)^4)-\ln (x(3e^{5x} +1))=\\ &=2x+4\ln (x^3-2)-\ln x-\ln (3e^{5x}+1).\end{align}$$ Now take derivative (implicitly) from both sides: $$\frac1y\cdot y'=2+\frac{12x^2}{x^3-2}-\frac1x-\frac{15e^{5x}}{3e^{5x}+1}$$ And the final answer is ...