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
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