- #1

- 65

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Could someone demonstrate the first step algebraically? Assuming c is the exponent on the variable expression, n is the numerator and d is the denominator, I tried:

[tex]c\frac{n(x)}{d(x)}\frac{n'(x)*d(x)-d'(x)*n(x)}{d(x)^2}[/tex]

Which gives me

[tex]2\frac{2x^2-1}{3x^4+2}\frac{[4x(3x^4+2)]-[(12x^3)(2x^2-1)]}{(3x^4+2)^2}[/tex]

Which simplifies to

[tex]\frac{-48 x^7+72 x^5+8 x^3-16 x}{(3x^4+2)^3}[/tex]

However, the book lists the answer as being [tex]\frac{-36x^5+24x^3+8x}{(3x^4+2)^2}[/tex]