Derivatives of Products and Quotients

Derivatives of Products and Quotients We have already seen that the derivative of a sum of functions is the sum of the derivatives, just as we saw with limits. We might be quick to jump to the conclusion that the derivative of a product of functions is also the product of the derivatives; this however, would be an incorrect conclusion. Product Rule for Derivatives If f (x) and g(x) are both differentiable, then dg df d (f (x) · g(x)) = f (x) · + g(x) · = f (x) · g 0 (x) + g(x) · f 0 (x) dx dx dx Thus, we have two terms, with each term containting the derivative of one of the two functions. The constant product rule is nothing but a special case of the product rule, where one of the functions is a constant. For such a function, its derivative is zero, so a single term remains, which is the derivative of the other function times the constant. Consider the following examples. Example 1 Find the derivative of p(x) = (x − 1)(x + 1). Solution Let f (x) = x − 1, and g(x) = x + 1. Since these are both linear functions with slope 1, it follows that f 0 (x) = 1 = g 0 (x). Now we can apply the product rule p0 (x) = f (x)g 0 (x) + g(x)f 0 (x) = (x − 1) · 1 + (x + 1) · 1 = 2x In this situation, we can also multiply the two polynomials out, and find that p(x) = x2 − 1 Applying the sum and product rules as we did in the last section we find p0 (x) = 2x + 0 = 2x which is consistent with the product rule. Example 2 Suppose the volume of a plant is described by V (t) = 100 + 12t where t is measured in days, and V is measured in cm3 . Let the density of the plant decrease according to ρ(t) = 0.8 − 0.05t where ρ is measured in grams per cm3 . Describe the change of plant’s mass over time. Solution First recall that mass is related to volume and density by M (t) = ρ(t)V (t) Now we can apply the product rule to find the rate of change of the mass M 0 (t) = ρ(t)V 0 (t)+ρ0 (t)V (t) = (0.8−0.05t)12+(100+12t)(−0.05) = 9.6−0.6t−5−0.6t = 4.6−1.2t Looking at the rate of change of mass as a function of time, we can see that initially the mass is increasing, but the rate it is increasing at decreases with time. Eventually, when t > 23 6 , the derivative becomes negative, so the mass begins to decrease. Similar to the product rule, we have a result that will allow us to calculate the derivative of a quotient. Quotient Rule for Derivatives If u(x) and v(x) are both differentiable, and v(x) 6= 0 then dv v(x) du v(x)u0 (x) − u(x)v 0 (x) d u(x) dx − dx u(x) = = dx v(x) v(x)2 v(x)2 This rule can be remembered as “low ‘d’ high minus high ‘d’ low,” which represents the numerator in the derivative of a quotient. ‘High’ is the numerator of the quotient, where ‘low’ is the denominator. Finally, ‘d’ represents the derivative of. 2 +1 . Example 3 Find the derivative of xx−1 2 Solution Let u(x) = x + 1, and v(x) = x − 1. It follows that u0 (x) = 2x and v 0 (x) = 1. Now we can use the quotient rule d u(x) v(x)u0 (x) − u(x)v 0 (x) (x − 1) · 2x − (x2 + 1) · 1 2x2 − 2x − x2 − 1 x2 − 2x − 1 = = = = dx v(x) v(x)2 (x − 1)2 (x − 1)2 (x − 1)2 Example 4 Find the derivative of the sinc function, which is commonly encountered in signal processing. Note that sin(x) sinc(x) = x Solution We will use the quotient rule and our knowledge that the derivative of the sine function is cosine to find x cos(x) − sin(x) d sinc(x) = dx x2 The set of Hill functions is an important family of functions in the study of biology, used to describe stimulus response. Hill functions take on the general form h(x) = xn 1 + xn where n ∈ Z+ (is a positive integer 1, 2, . . .). Example 5 Calculate the derivative of the Hill functions for n = 1 and n = 2. Solution We will begin with n = 1. Let u(x) = x and v(x) = 1+x. It follows that u0 (x) = v 0 (x) = 1. Using the quotient rule we find (1 + x) · 1 − x · 1 1 dh = = 2 dx (x + 1) (x + 1)2 This derivative is always positive for x > 0, and h0 (0) = 1. Now let n = 1, so u(x) = x2 and v(x) = 1 + x2 . It follows that u0 (x) = v 0 (x) = 2x. From the quotient rule we find (1 + x2 ) · 2x − x2 · 2x 2x dh = = 2 dx (x2 + 1)2 (x + 1)2 This derivative is also always positive for x > 0, but has the value h0 (0) = 0.