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Simplify the derivative by applying the properties of logarithms
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$\frac{d}{dx}\left(\frac{\frac{x^3-1}{x^3-2x^2-3x}\frac{x+1}{\left(x^2+x-2\right)\left(x^2+x+1\right)}}{6x+x^2-x^3}\right)$
Learn how to solve problems step by step online. Find the derivative using logarithmic differentiation method ((x^3-1)/(x^3-2x^2-3x)((x+1)/(x^2+x+-2))/(x^2+x+1))/(6x+x^2-x^3). Simplify the derivative by applying the properties of logarithms. Multiplying fractions \frac{x^3-1}{x^3-2x^2-3x} \times \frac{x+1}{\left(x^2+x-2\right)\left(x^2+x+1\right)}. Divide fractions \frac{\frac{\left(x^3-1\right)\left(x+1\right)}{\left(x^3-2x^2-3x\right)\left(x^2+x-2\right)\left(x^2+x+1\right)}}{6x+x^2-x^3} with Keep, Change, Flip: \frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}. To derive the function \frac{\left(x^3-1\right)\left(x+1\right)}{\left(x^3-2x^2-3x\right)\left(x^2+x-2\right)\left(x^2+x+1\right)\left(6x+x^2-x^3\right)}, use the method of logarithmic differentiation. First, assign the function to y, then take the natural logarithm of both sides of the equation.