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Example Find the derivative of the following function．

$\displaystyle \displaystyle{y = x^3 \sin{3x}} $

Answer To find the derivative of a product, it is important to know the product 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. Product Rule． $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$  
4. Basic Formula for Derivatives. $(\tan{x})' = \sec^{2}{x}, (\sec{x})' = \sec{x}\tan{x}$  

Apply these rules to the above problem, we find

$\displaystyle (x^3 \sin{3x})'$ $\displaystyle =$ $\displaystyle (x^3)'\sin{3x} + x^3(\sin{3x})'$  
  $\displaystyle =$ $\displaystyle 3x^2 \sin{3x} + 3x^3\cos{3x}$