11.4. Differentiating ProductsandQuotients. Introduction. Prerequisites. Learning Outcomes

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1 Differentiating ProductsandQuotients 11.4 Introduction We have seen, in the first three Sections, how standard functions like n, e a, sin a, cos a, ln a may be differentiated. In this Section we see how more complicated functions may be differentiated. We concentrate, for the moment, on products and quotients of standard functions like n e a, ea ln sin. We will see that two simple rules may be consistently employed to obtain the derivatives of such functions. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... HELM (2008): be able to differentiate the standard functions: logarithms, polynomials, eponentials, and trigonometric functions be able to manipulate algebraic epressions differentiate products and quotients of the standard functions differentiate a quotient using the product rule 29

2 1. Differentiating a product In previous Sections we have eamined the process of differentiating functions. We found how to obtain the derivative of many commonly occurring functions. These are recorded in the following table (remember, arguments of trigonometric functions are assumed to be in radians). Table 2 y n sin a cos a tan a sec a ln a e a cosh a sinh a n n 1 a cos a a sin a a sec 2 a a sec tan 1 ae a a sinh a a cosh a In this Section we consider how to differentiate non-standard functions - in particular those which can be written as the product of standard functions. Being able to differentiate such functions depends upon the following Key Point. If y = f()g() then Key Point 9 Product Rule = g() + f() (or y = f g + fg ) If y = u.v then = udv + v du (or y = uv + vu ) These versions are equivalent, and called the product rule. We shall not prove this result, instead we shall concentrate on its use. 30 HELM (2008): Workbook 11: Differentiation

3 Eample 8 Differentiate (a) y = 2 sin (b) y = ln Solution (a) Here f() = 2, g() = sin and so = 2, = cos = 2(sin ) + 2 (cos ) = (2 sin + cos ) (b) Here f() =, g() = ln and so = 1.(ln ) +. = 1, ( ) 1 = ln + 1 = 1 Task Determine the derivatives of the following functions (a) y = e ln, (b) y = e2 2 (a) Use the product rule: f() = = = g() = = = e ln + e (b) Write y = ( 2 )e 2 and then differentiate: f() = = = g() = = = ( 2 3 )e (2e 2 ) = 2e2 ( 1 + ) 3 HELM (2008): 31

4 The rule for differentiating a product can be etended to any number of products. If, for eample, y = f()g()h() then = d [g()h()] + f() [g()h()] = g()h() + f() { h() + g()dh = g()h() + f() h() + f()g()dh That is, each function in the product is differentiated in turn and the three results added together. } Eample 9 If y = e 2 sin then find. Solution Here f() =, g() = e 2, h() = sin = 1, = 2e2, dh = cos = 1(e2 sin ) + (2e 2 ) sin + e 2 (cos ) = e 2 (sin + 2 sin + cos ) Task Obtain the first derivative of y = 2 (ln ) sinh. Firstly identify the three functions: f() = g() = h() = f() = 2, g() = ln, h() = sinh Now find the derivative of each of these functions: 32 HELM (2008): Workbook 11: Differentiation

5 = = dh = = 2, = 1, dh = cosh Finally obtain : ( ) 1 = 2(ln ) sinh + 2 sinh + 2 ln (cosh ) = 2 ln sinh + sinh + 2 ln cosh Task Find the second derivative of y = 2 (ln ) sinh by differentiating each of the three terms making up found in the previous Task (2 ln sinh, sinh, 2 ln cosh ), and finally, simplify your answer by collecting like terms together: d (2 ln sinh ) = d ( sinh ) = d (2 ln cosh ) = d 2 y 2 = d 2 y 2 = (2 + 2 ) ln sinh + 3 sinh + 2 cosh + 4 ln cosh HELM (2008): 33

6 Eercises 1. In each case find the derivative of the function (a) y = tan (b) y = 4 ln(2) (c) y = sin 2 (d) y = e 2 cos 3 2. Find the derivatives of: (a) y = cos (b) y = e sin s (c) Obtain the derivative of y = e tan using the results of parts (a) and (b). 1. (a) = tan + sec2 (b) = 43 ln(2) + 4 = 3 (4 ln(2) + 1) (c) y = sin. sin = cos sin + sin cos = 2 sin cos = sin 2 (d) = (2e2 ) cos 3 + e 2 ( 3 sin 3) = e 2 (2 cos 3 3 sin 3) 2. (a) y = sec = sec + sec tan (b) = e sin + e cos = e (sin + cos ) (c) The derivative of y = e tan = ( sec )(e sin ) is found by applying the product rule to the results of (a) and (b): = d ( sec ).e sin + ( sec ) d (e sin ) = (sec + sec tan )e sin + sec (e )(sin + cos ) = e ( + tan + tan + tan 2 ) 34 HELM (2008): Workbook 11: Differentiation

7 2. Differentiating a quotient In this Section we consider functions of the form y = f(). To find the derivative of such a function g() we make use of the following Key Point: Key Point 10 Quotient Rule If y = f() g() then g() = f() [g()] 2 (or y = gf g f g 2 ) If y = u v then du v = dv u v 2 (or y = vu v u v 2 ) These are two equivalent versions of the quotient rule. Eample 10 Find the derivative of y = ln Solution Here f() = ln and g() = = 1 and = 1 ( ) 1 Hence 1(ln ) = = 1 ln [] 2 2 HELM (2008): 35

8 Task Obtain the derivative of y = sin 2 (a) using the formula for differentiating a product and (b) using the formula for differentiating a quotient. (a) Write y = 2 sin then use the product rule to find : y = 2 sin = ( 2 3 ) sin + 2 cos = 2 sin + cos 3 (b) Now use the quotient rule instead to find : y = sin 2 = 2 (cos ) (2) sin = ( 2 ]) 2 cos 2 sin 3 36 HELM (2008): Workbook 11: Differentiation

9 Eercise Find the derivatives of the following: (a) ( )( ) (b) (c) (d) ( 2 + 3)(2 5)(3 + 2) (2 + 1)(3 1) (e) + 5 (f) (ln ) sin (g) (ln )/ sin (h) e / 2 (i) e sin cos 2 (a) (b) ( ) 2 4( + 1) (c) ( 1) 3 (d) (e) 6( ) ( + 5) 2 (f) 1 sin + (ln ) cos ( ) 1 sin (ln ) cos (g) sin 2 = cosec( 1 cot ln ) (h) 2 e 2e = ( )e 4 (i) cos 2(e sin + e cos ) + 2 sin 2e sin cos 2 2 = e [(sin + cos ) sec sin sin 2 sec 2 2] HELM (2008): 37