20. Product rule, Quotient rule Product rule, Quotient rule

Product rule 20.1. Product rule Quotient rule

We have seen that the derivative of a sum is the sum of the derivatives: d d d [f(x) + g(x)] = [f(x)] + [(g(x)] . dx dx dx One might expect from this that the derivative of a product is the product of the derivatives. This is not the case, however. In fact, it usually happens that d d d [f(x)g(x)] 6= [f(x)] [g(x)] . Table of Contents dx dx dx For instance, d d d d JJ II [xx] = x2 = 2x 6= 1 = (1)(1) = [x] [x] . dx dx dx dx J I Instead, the rule for finding the derivative of a product is as follows:

Page1 of 10 Product rule. For functions f and g, Back d d d [f(x)g(x)] = [f(x)] g(x) + f(x) [g(x)] . dx dx dx Print Version

In words, the derivative of a product is the derivative of the first times the second plus the Home Page first times the derivative of the second. For example, Product rule, Quotient rule d d d x3 sin x = x3 sin x + x3 [sin x] Product rule dx dx dx Quotient rule = 3x2 sin x + x3 cos x.

With p(x) = f(x)g(x), the rule says that p0(x) = f 0(x)g(x) + f(x)g0(x), so we verify the rule by establishing this equation using the definition of the derivative: p(x + h) − p(x) p0(x) = lim h→0 h f(x + h)g(x + h) − f(x)g(x) = lim Table of Contents h→0 h f(x + h)g(x + h) − f(x)g(x + h) + f(x)g(x + h) − f(x)g(x) = lim h→0 h JJ II f(x + h) − f(x) g(x + h) − g(x) = lim g(x + h) + f(x) h→0 h h J I  f(x + h) − f(x)  g(x + h) − g(x) = lim lim g(x + h) + f(x) lim h→0 h h→0 h→0 h Page2 of 10 = f 0(x)g(x) + f(x)g0(x).

20.1.1 Example Find the derivatives of the following functions: Back

(a) f(x) = (x8 + 2x − 3)ex. Print Version

t 2 (b) f(t) = 5 cos t + 4t . Home Page

Solution (a) Product rule, Quotient rule d f 0(x) = (x8 + 2x − 3)ex Product rule dx Quotient rule d d = x8 + 2x − 3 ex + (x8 + 2x − 3) [ex] dx dx = (8x7 + 2)ex + (x8 + 2x − 3)ex = (x8 + 8x7 + 2x − 1)ex.

(b) Here, we need to use the sum rule before using the product rule: d f 0(t) = 5t cos t + 4t2 dt Table of Contents d d = 5t cos t + 4t2 dt dt d d JJ II = 5t cos t + 5t [cos t] + 8t dt dt = (5t ln 5) cos t + 5t(− sin t) + 8t J I = 5t(ln 5) cos t − 5t sin t + 8t. Page3 of 10

Back The product rule extends naturally to handle any number of factors. For instance, Print Version d [f(x)g(x)h(x)] = dx Home Page d d d [f(x)]g(x)h(x) + f(x) [g(x)] h(x) + f(x)g(x) [h(x)] . dx dx dx The derivative is obtained by taking the derivative of one factor at a time, leaving the Product rule, Quotient rule other factors unchanged, and then summing the results. This rule is verified by using the product rule repeatedly (see Exercise 20–3). Product rule Quotient rule 20.1.2 Example Find the derivative of f(x) = (x3 − 4x2)ex cos x.

Solution d f 0(x) = (x3 − 4x2)ex cos x dx d d = x3 − 4x2 ex cos x + (x3 − 4x2) [ex] cos x dx dx d Table of Contents + (x3 − 4x2)ex [cos x] dx = (3x2 − 8x)ex cos x + (x3 − 4x2)ex cos x + (x3 − 4x2)ex(− sin x) JJ II = (x3 − x2 − 8x)ex cos x − (x3 − 4x2)ex sin x.

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Back Next, we get the rule for finding the derivative of a quotient.

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Home Page Product rule, Quotient rule Quotient rule. For functions f and g, Product rule d d Quotient rule d f(x) g(x) [f(x)] − f(x) [g(x)] = dx dx . dx g(x) (g(x))2

In words, the derivative of a quotient is the bottom times the derivative of the top minus the top times the derivative of the bottom, over the bottom squared. The verification (omitted) is very similar to that for the product rule. Table of Contents 20.2.1 Example Find the derivatives of the following functions:

x4 − 2x3 + 8 JJ II (a) f(x) = . x7 − x 3 sin t J I (b) f(t) = . t2 − et Page5 of 10 Solution

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Home Page (a) Product rule, Quotient rule

d x4 − 2x3 + 8 Product rule f 0(x) = dx x7 − x Quotient rule d d (x7 − x) x4 − 2x3 + 8 − (x4 − 2x3 + 8) x7 − x = dx dx (x7 − x)2 (x7 − x)(4x3 − 6x2) − (x4 − 2x3 + 8)(7x6 − 1) = . (x7 − x)2

(b) Table of Contents d  3 sin t  f 0(t) = dt t2 − et d d JJ II (t2 − et) [3 sin t] − (3 sin t) t2 − et = dt dt (t2 − et)2 J I (t2 − et)(3 cos t) − (3 sin t)(2t − et) = 2 t 2 . (t − e ) Page6 of 10

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Sometimes a quotient to be differentiated can be rewritten in such a way that the quotient Print Version rule becomes unnecessary. In this case, going ahead and rewriting is usually preferable to using the quotient rule; the quotient rule should be used only if it cannot be avoided. Home Page 20.2.2 Example Find the derivative of the function Product rule, Quotient rule √ x3 − 5x + 4 x Product rule f(x) = . x Quotient rule Solution At the appropriate step, the function is rewritten in order to avoid using the quotient rule: √ d x3 − 5x + 4 x f 0(x) = dx x d h i = x2 − 5 + 4x−1/2 dx = 2x − 2x−3/2 Table of Contents 2 = 2x − √ . ( x)3 JJ II

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Home Page 20 – Exercises Product rule, Quotient rule

Product rule Quotient rule

20 – 1 Let f(x) = (x2 − 3x + 1)(x4 + 9x2). (a) Find the derivative of f by first expanding the right-hand side so as to avoid using the product rule. (b) Find the derivative of f by using the product rule and verify that the result is the same as that obtained in part (a).

Table of Contents 20 – 2 Find the derivatives of the following functions: (a) f(x) = (2x5 − 3x2 + 1) sin x JJ II √ 5 (b) f(t) = 3 tet − J I t

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Home Page 20 – 3 Verify the formula Product rule, Quotient rule

d Product rule [f(x)g(x)h(x)] = dx Quotient rule d d d [f(x)]g(x)h(x) + f(x) [g(x)] h(x) + f(x)g(x) [h(x)] . dx dx dx

Hint: Apply the product rule viewing f(x)g(x)h(x) as a product with the two factors f(x) and g(x)h(x).

√ 20 – 4 Find the derivative of f(x) = (2x − 5 + 3 x)8x cos x. Table of Contents Hint: Use the formula in Exercise 20–3.

√ JJ II x4 + 3 x − 5 20 – 5 Let f(x) = . x J I (a) Find the derivative of f by first rewriting the right-hand side so as to avoid using the quotient rule. Page9 of 10 (b) Find the derivative of f by using the quotient rule and verify that the result is the same as that obtained in part (a). Back

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Home Page 20 – 6 Find the derivatives of the following functions: Product rule, Quotient rule 3x2 (a) f(x) = Product rule 5x − 7 Quotient rule 4 cos t − 1 (b) f(t) = 2 + 3et

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