Do We Need the Quotient Rule? The quotient rule can be diﬃcult to memorize, and some students are more comfortable with negative exponents than they are with fractions. In this exer cise we learn how we can use the chain and product rules together in place of the quotient rule. a) Use the quotient rule to ﬁnd the derivative of x3 . x+1 b) Use the product and chain rules to ﬁnd the derivative of x3 · (x + 1)−1 . Note x3 that x3 · (x + 1)−1 = . x+1 c) Use the chain and product rules (and not the quotient rule) to show that the u� (x)v(x) − u(x)v � (x) derivative of u(x)(v(x))−1 equals . (v(x))2 Solution a) Use the quotient rule to ﬁnd the derivative of x3 . x+1 The quotient rule tells us that: � �� u(x) u� (x)v(x) − u(x)v � (x) = . v(x) (v(x))2 Here u(x) = x3 , u� (x) = 3x2 , v(x) = x + 1 and v � (x) = 1. Therefore, � 3 �� x 3x2 (x + 1) − x3 · 1 = x+1 (x + 1)2 3x3 + 3x2 − x3 = (x + 1)2 3 2x + 3x2 = (x + 1)2 b) Use the product and chain rules to ﬁnd the derivative of x3 · (x + 1)−1 . (Note x3 that x3 · (x + 1)−1 = .) x+1 The product rule tells us that (u(x) · v(x))� = u� (x)v(x) + u(x)v � (x). Once again u(x) = x3 and u� (x) = 3x2 , but now v(x) = (x + 1)−1 . We apply the chain rule to ﬁnd v � (x) = −1 · (x + 1)−2 · 1 = −(x + 1)−2 . Now we can apply the product rule to ﬁnd that: (x3 · (x + 1)−1 )� = 3x2 (x + 1)−1 + x3 (−(x + 1)−2 ) = 3x2 (x + 1)−1 − x3 (x + 1)−2 1 x3 , the derivatives of the two expressions look x+1 very diﬀerent. In fact, they are algebraically equivalent; diﬀerent-looking but equivalent answers are a common occurrence in calculus. We can show the equivalence of the two answers by applying some basic algebra: Although x3 · (x + 1)−1 = 3x2 (x + 1)−1 − x3 (x + 1)−2 = = = = = 3x2 (x + 1)−1 · ((x + 1)−1 · (x + 1)+1 ) − x3 (x + 1)−2 create a common factor of (x + 1)−2 2 3x (x + 1)−2 · (x + 1) − x3 (x + 1)−2 factor out (x + 1)−2 (3x2 (x + 1) − x3 )(x + 1)−2 eliminate the negative exponent. 2 3x (x + 1) − x3 (x + 1)2 3 2x + 3x2 (x + 1)2 c) Use the chain and product rules (and not the quotient rule) to show that the f � (x)g(x) − f (x)g � (x) . derivative of f (x)(g(x))−1 equals (g(x))2 We can use the previous solution as an outline for this one. Here u(x) = f (x), u� (x) = f � (x), v(x) = (g(x))−1 and, by the chain rule, v � (x) = −(g(x))−2 g � (x). Hence, (f (x)(g(x))−1 )� = f � (x)(g(x))−1 + f (x)(−(g(x))−2 g � (x)) = f � (x)(g(x))−1 − f (x)(g(x))−2 g � (x) = f � (x)(g(x))−2 g(x) − f (x)g � (x)(g(x))−2 = [f � (x)g(x) − f (x)g � (x)](g(x))−2 = f � (x)g(x) − f (x)g � (x) (g(x))2 Note that if time permits, you can use this alternate method of diﬀerentiation to check your work on problems involving the chain rule.