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Example Find the derivative of the following function．

$\displaystyle \displaystyle{y = \frac{x^2+1}{x^3 -1}} $

Answer To find the derivative of a quotient, it is important to knou the quotient rule.

1. The derivative of a sum is the sum of the derivatives． $(f(x) + g(x))' = f'(x) + g'(x)$  
2. The derivative of a constant multiple is the constant times the derivative． $(\alpha f(x))' = \alpha f'(x)$  
3. The product rule． $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$  
4. The quotient rule. $(\frac{f(x)}{g(x)})' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$  
5. Basic formula for derivatives. $(\tan{x})' = \sec^{2}{x}, (\sec{x})' = \sec{x}\tan{x}$  

Apply these rules to the above problem, we have

$\displaystyle (\frac{x^2+1}{x^3 -1})'$ $\displaystyle =$ $\displaystyle \frac{(x^2+1)'(x^3 - 1) - (x^2+1)(x^3 -1)'}{(x^3 -1)^2}$  
  $\displaystyle =$ $\displaystyle \frac{2x(x^3 -1) -(x^2+1)(3x^2)}{(x^3 -1)^2}$  
  $\displaystyle =$ $\displaystyle \frac{-x^4 -3x^2-2x}{(x^3 -1)^2}$  
  $\displaystyle =$ $\displaystyle \frac{-x(x^3 + 3x + 2)}{(x^3 -1)^2}$