Find the derivative of $\frac{\frac{x^3-1}{x^3-2x^2-3x}\frac{\frac{x+1}{x^2+x-2}}{x^2+x+1}}{6x+x^2-x^3}$

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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 simplify trigonometric expressions problems step by step online. Find the derivative of ((x^3-1)/(x^3-2x^2-3x)((x+1)/(x^2+x+-2))/(x^2+x+1))/(6x+x^2-x^3). Simplifying. 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}. Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}.