Find conditions on a, b, c, and d such that commutes with every 2 2 matrix.

Product Rule If F(x)=f(x)g(x), Then F’(x)=f(x)g’(x) + f’(x)g(x) Ex. d/dx [ (4x^3 - x^2 -1)(x^3 -2x^2 + 3x +1)] = (4x^3 - X^2 - 1) (3x^2- 4x + 3) = (12x^2 -2x)(x^3 - 2x^2 + 3x + 1) However the product rule is more useful for the product of different types of functions Ex. f(x)= x^2cos(x) g(x) = e^x arctan(x) Proof of the Product Rule Let F(x) = f(x)g(x) F’(x) = lim ((F(x + h ) - F(x))/h) h 0 =lim ((f(x + h )g( x + h) - f(x)g(x))/h) h 0 Subtract and add f(x + h)g(x) = lim ((f( x + h)g(x + h) - f(x + h)g(x) + f(x + h) - g(x) - f(x)g(x))/h) h 0 = lim ((f( x + h)(g( x + h) - g(x)) + g(x)(f( x + h) - f(x))/h) h 0 = lim ((f ( x+ h)(g( x + h) - g(x))/h) + (g(x)f(x + h) - f(x)/h) h 0 = lim (f( x + h) lim ((g(x + h) - g(x))/h) + lim (g(x))lim((f(x + h) - f(x))/h) h 0 h 0 h 0 h 0 = f(x)g’(x) + g(x)f’(x) The Quotient Rule d/dx (f(x)/g(x)) = g(x)d/dx f(x) - f(x)d/dx (g(x))/(g(x))^2 If F(x) = f(x)/g( ’(x) = (g(x)f’(x) - f(x)g’(x))/(g(x)^2) Rhyme for remembering quotient rule “ Low times d-high minus high. D-low, over the square of what’s below” Ex. d/dx [x/x^2-1] = ((x^2-1)(1) - x(2x))/(x^2-1)^2 = (x^2 - 1 - 2x^2)/ ( x^4 - 2x^2 + 1) = (- x^2 - 1)/ ( X^4 - 2x^2 + 1) d/dx( x^2 + 2x)/(x^4 - 3x^2 + 1) = ((x^4 - 3x^2 +1)(2x + 2) - (x^2 + 2x)(4x^3 - 6x))/(x^4 - 3x^2 + 1)^2 Graph Note. with a function like f(x) = 3/ = 3/x^½ It is easier to rewrite it a 3x^-½ and use the power rule =f’(x) = -(3/2)x^-3/2 Using the quotient rule F’(x) = (x^(½