“Teach A Level Maths” Vol. 2: A2 Core Modules 12: The Quotient Rule © Christine Crisp The Quotient Rule Module C3 AQA Edexcel MEI/OCR "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" The Quotient Rule The following are examples of quotients: x x (a) y sin x e (b) y 1 x 1 x2 sin x (d) y cos x (c) y x2 (c) can be divided out to form a simple function as there is a single polynomial term in the denominator. y 1 x2 x 2 y 1 2 x2 x x2 dy 2 y x 1 2 x 3 dx For the others we use the quotient rule. The Quotient Rule The quotient rule gives us a way of differentiating functions which are divided. The rule is similar to the product rule. u y v du dv v u dy dx 2 dx dx v where u and v are functions of x. This rule can be derived from the product rule but it is complicated. If you want to go straight to the examples, click on the box below. Examples The Quotient Rule We can develop the quotient rule by using the product rule! u y v 1 yu v y u v 1 The problem now is that this v is not the same as the v of the product rule. That v is replaced by v 1. 1 dy du dv dy du d ( v ) 1 So, becomes v u v u dx dx dx dx dx dx 1 dy 1 du d ( v ) Simplifying u dx v dx dx 1 d ( v ) , is the derivative of v 1 Part of the term, dx but with respect to x not v. 2nd The Quotient Rule 1 d ( v ) We use the chain rule: dx 1 d ( v ) So, dx 1 dy 1 du d ( v ) Then, u dx v dx dx d (v 1 ) dv dv dx 1 dv 2 dx v dy 1 du u dv 2 dx v dx v dx Make the denominators the same by multiplying the numerator and denominator of the 1st term by v. dy v du u dv 2 2 du dv dx v dx v dx v u dy Write with a common denominator: dx 2 dx dx v The Quotient Rule x2 e.g. 1 Differentiate y to find dy . x 1 dx u Solution: y u x 2 and v x 1 v dv du 1 2x dx dx du dv v u u dy y dx 2 dx v dx v dy 2 x( x 1) x 2 dx ( x 1) 2 We now need to simplify. The Quotient Rule dy 2 x( x 1) x 2 dx ( x 1) 2 We could simplify the numerator by taking out the common factor x, but it’s easier to multiply out the brackets. We don’t touch the denominator. 2 2 dy 2 x 2 x x Multiplying out numerator: dx ( x 1) 2 dy x2 2x Now collect like terms: dx ( x 1) 2 dy x ( x 2) and factorise: dx ( x 1) 2 We leave the brackets in the denominator as the factorised form is simpler. The Quotient Rule Quotients can always be turned into products. e.g. ex 1 x 2 can be written as e x (1 x 2 ) 1 However, differentiation is usually more awkward if we do this. In the quotient above, u e x and v 1 x 2 ( both simple functions ) In the product , u e x and v (1 x 2 ) 1 ( v needs the chain rule ) SUMMARY The Quotient Rule To differentiate a quotient: Check if it is possible to divide out. If so, do it and differentiate each term. Otherwise use the quotient rule: If u y v , where u and v are both functions of x du dv v u dy dx 2 dx dx v The Quotient Rule Exercise Use the quotient rule, where appropriate, to differentiate the following. Try to simplify your answers: 1. y x3 e x 2. y 4. y 2 3. x y 2 x cos x x4 2 x x2 The Quotient Rule 1. y x3 e Solution: x u y v du dv v u dy dx 2 dx dx v u y u x3 v du 3x 2 dx and v ex dv ex dx dy 3 x 2 e x x 3 e x x 2 dx e 2 dy x 2 e x ( 3 x ) dy x ( 3 x ) 2 x x dx dx e e 2. y cos x x Solution: The Quotient Rule 4 u y v du dv v u dy dx 2 dx dx v u y u cos x and v x 4 v dv du 3 4x sin x dx dx dy x 4 sin x 4 x 3 cos x 4 2 dx x ( x sin x 4 cos x ) dy x 3 ( x sin x 4 cos x ) dx x5 x8 5 x The Quotient Rule 3. x2 y 2 x Solution: u y v du dv v u dy dx 2 dx dx v u y u x 2 and v 2 x v dv du 1 2x dx dx 2 dy 2 x( 2 x ) x ( 1) 2 dx (2 x ) 2 2 2 dy 4 x 2 x x 4x x x(4 x ) 2 2 dx (2 x ) (2 x ) (2 x ) 2 4. y The Quotient Rule 2 x x2 Solution: y Divide out: 2 x x 2 y 2 2 1x x x2 y 2 x 2 x 1 dy 3 2 4 x x dx 4 1 3 2 x x The Quotient Rule We can now differentiate the trig function by writing y tan x sin x y tan x cos x u v cos x u sin x y v dv du sin x cos x dx dx dy (cos x )(cos x ) (sin x )( sin x ) dx (cos x )(cos x ) cos2 x sin2 x cos2 x The Quotient Rule So, y tan x dy cos2 x sin2 x dx cos2 x This answer can be simplified: dy 1 cos x sin x 1 dx cos2 x 1 Also, is defined as secx cos x 2 So, 2 y tan x dy sec2 x dx The Quotient Rule Those of you taking the OCR/MEI spec can omit the next ( final exercise ). The Quotient Rule Exercise Use the quotient rule ( or, for (a) and (b), the chain rule ) to find the derivatives with respect to x of (a) y cosec x (b) y sec x (c) y cot x Before you check the solutions, look in your formula books to see the forms used for the answers. Try to get your answers into these forms. The Quotient Rule (a) Solution: 1 y cosec x sin x dy sinx(0) 1(cos x ) dx (sinx ) 2 cos x sin x sin x cos x 1 sin x sin x cot x cosec x ( cose cx cot x ) The Quotient Rule (b) Solution: 1 y sec x cos x dy cos x(0) 1( sinx ) 2 dx (cos x ) sin x cos x cos x sin x 1 cos x cos x tan x sec x ( se cx tan x ) The Quotient Rule (c) Solution: cos x y cot x sin x dy sin x( sin x ) cos x cos x 2 dx (sinx ) (sin2 x cos2 x ) sin2 x 1 2 si n x cosec2 x The Quotient Rule The Quotient Rule The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet. The Quotient Rule The following are examples of quotients: x x (a) y sin x e (b) y 1 x 1 x2 sin x (d) y cos x (c) y x2 (c) can be divided out to form a simple function as there is a single polynomial term in the denominator. y 1 x2 x 2 y 1 2 x2 x x2 dy 2 y x 1 2 x 3 dx For the others we use the quotient rule. The Quotient Rule SUMMARY To differentiate a quotient: Check if it is possible to divide out. If so, do it and differentiate each term. Otherwise use the quotient rule: If u y v , where u and v are both functions of x du dv v u dy dx 2 dx dx v The Quotient Rule x2 e.g. 1 Differentiate y to find dy . x 1 dx u Solution: y u x 2 and v x 1 v dv du 1 2x dx dx du dv v u u dy y dx 2 dx v dx v dy 2 x( x 1) x 2 dx ( x 1) 2 We now need to simplify. The Quotient Rule dy 2 x( x 1) x 2 dx ( x 1) 2 We could simplify the numerator by taking out the common factor x, but it’s easier to multiply out the brackets. We don’t touch the denominator. 2 2 dy 2 x 2 x x Multiplying out numerator: dx ( x 1) 2 dy x2 2x Now collect like terms: dx ( x 1) 2 dy x ( x 2) and factorise: dx ( x 1) 2 We leave the brackets in the denominator. ( A factorised form is considered to be simpler. ) The Quotient Rule We can now differentiate the trig function by writing y tan x sin x y tan x cos x u v cos x u sin x y v dv du sin x cos x dx dx dy (cos x )(cos x ) (sin x )( sin x ) dx (cos x )(cos x ) cos2 x sin2 x cos2 x The Quotient Rule So, y tan x dy cos2 x sin2 x dx cos2 x This answer can be simplified: dy 1 cos x sin x 1 dx cos2 x 1 Also, is defined as secx cos x 2 So, 2 y tan x dy sec2 x dx