Prove the Quotient Rule using the Product Rule and Chain Rule

given that the chain rule is d/dx(f(g(x))) = g'(x)f'(g(x))given that the product rule is d/dx(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)given that the quotient rule is d/dx(f(x)/g(x)) = (g(x)f'(x) - g'(x)f(x))/(g(x))²
LHS:d/dx(f(x)/g(x)) = d/dx(f(x)(g(x))^-1)
let h(x) = (g(x))^-1
by using the chain ruleh'(x) = -g'(x)(g(x))^-2
therefor: LHS = d/dx(f(x)h(x))
by using the product ruleLHS = f'(x)h(x) + f(x)h'(x)
by substituting the values of h(x) and h'(x)LHS = f'(x)(g(x))^-1 - f(x)g'(x)(g(x))^-2
by rearranging and turning into a fraction with a denominator of (g(x))²LHS = (g(x)f'(x) - g'(x)f(x))/(g(x))² = RHSas required