No, because d/dx (1/g(x)) is not 1/g'(x). You have to apply the chain rule. It should be -1/(g(x))² · g'(x) So we'd have (using the product rule and the chain rule): d/dx f(x) · 1/g(x) = 1/g(x) · f'(x) + f(x) · -1/(g(x))² · g'(x) Which, with a bit of manipulation, can be made to look like the familiar quotient rule for differentiation.
1(x) = x. The first few are P 2(x) = 3 2 (x 2 −1 3), P 3(x) = 5 2 (x 3 −3 5 x) and P 4(x) = 1 8 (35x 4 −30x2 +3). Each P k(x) is a polynomial of degree k, and has k roots that all lie in the interval (−1,1). So, to find the quadrature rule with maximum degree of exactness using n nodes and n weights, in principle we need to: •Find