# Calculate the derivative of the following function f(x)-\ln\frac{(2x-1)(x+2)^3}{(1-4x)^2}

Calculate the derivative of the following function
$f\left(x\right)-\mathrm{ln}\frac{\left(2x-1\right){\left(x+2\right)}^{3}}{{\left(1-4x\right)}^{2}}$
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otoplilp1
$f\left(x\right)-\mathrm{ln}\frac{\left(2x-1\right){\left(x+2\right)}^{3}}{{\left(1-4x\right)}^{2}}$
$f\left(x\right)=\mathrm{ln}\left(\left(2x-1\right){\left(x+2\right)}^{3}\right)-{\mathrm{ln}\left(1-4x\right)}^{2}\left[\because \mathrm{ln}\left(\frac{a}{b}\right)=\mathrm{ln}a-\mathrm{ln}b\right]$
$f\left(x\right)=\mathrm{ln}\left(2x-1\right)+{\mathrm{ln}\left(x+2\right)}^{3}-{\mathrm{ln}\left(1-4x\right)}^{2}\left[\because \mathrm{ln}\left(ab\right)=\mathrm{ln}a+\mathrm{ln}b\right]$
$f\left(x\right)=\mathrm{ln}\left(2x-1\right)+3\mathrm{ln}\left(x+2\right)-2\mathrm{ln}\left(1-4x\right)\left[\because \mathrm{ln}\left({x}^{n}\right)=n\mathrm{ln}x\right]$
${f}^{\prime }\left(x\right)=\frac{d}{dx}\mathrm{ln}\left(2x-1\right)+3\frac{d}{dx}\mathrm{ln}\left(x+2\right)-2\frac{d}{dx}\mathrm{ln}\left(1-4x\right)$
$=\frac{1}{2x-1}\frac{d}{dx}\left(2x-1\right)+\frac{3}{x+2}\frac{d}{dx}\left(x+2\right)-\frac{2}{1-4x}\frac{d}{dx}\left(1-4x\right)$
$\left[\because \frac{d}{dx}\mathrm{ln}x=\frac{1}{x}\right]$
$=\frac{1}{2x-1}\left(2\cdot 1-0\right)+\frac{3}{x+2}\left(1+0\right)-\frac{2}{1-4x}\left(0-4\cdot 1\right)\left[\because \frac{d}{dx}x=1\right]$
Heather Fulton
Is ${f}^{\prime }\left(x\right)=\frac{2}{2x-1}+\frac{3}{x+2}+\frac{8}{1-4x}$ final answer?