Differentiation

1 ex tanx+sec2x B1 Finds first derivative. ( ) ex tanx+2sec2x+2sec2xtanx M1 A1 Finds second derivative. y(0)=0, y'(0)=1, y''(0)=2 M1 Evaluates derivatives at x=0. y=x+x2 A1 5 Question Answer Marks Guidance

3(a) 3xy2y'+ y3 B1 Differentiates xy3. ( ) −4 x3y'+3x2y =0 B1 Differentiates x3y. ( ) ( ) 3(−1)(1)y'+13−4 (−1)3y'+3(−1)2(1) =0 leading to y'=11 B1 Substitutes (−1,1), AG. y' 3xy2 −4x3 =12x2y− y3 3 © UCLES 2021 Page 6 of 13 3(b) 3xy2 −4x3 y''+ y' 6xyy'+3y2 −12x2 B1 B1 Differentiates 3xy2 −4x3 y'. ( ) +3y2y'−12 x2y'+2xy =0 B1 Differentiates y3−12x2y. 3xy2y''−4x3y''+6xy ( y' )2 +6y2y'−24x2y'−24xy=0 ( )( ) ( )( ) 3xy2 −4x3 12x2y'+24xy−3y2y' − 12x2y− y3 6xyy'+3y2 −12x2 y''= ( 3xy2 −4x3 )2 y''+11(−75)+3(11)−12(9)=0 M1 Substitutes (–1, 1) y''=900 A1 5 © UCLES 2021 Page 7 of 13

1 ex tanx+sec2x B1 Finds first derivative. ( ) ex tanx+2sec2x+2sec2xtanx M1 A1 Finds second derivative. y(0)=0, y'(0)=1, y''(0)=2 M1 Evaluates derivatives at x=0. y=x+x2 A1 5 Question Answer Marks Guidance

3(a) 3xy2y'+ y3 B1 Differentiates xy3. ( ) −4 x3y'+3x2y =0 B1 Differentiates x3y. ( ) ( ) 3(−1)(1)y'+13−4 (−1)3y'+3(−1)2(1) =0 leading to y'=11 B1 Substitutes (−1,1), AG. y' 3xy2 −4x3 =12x2y− y3 3 © UCLES 2021 Page 6 of 13 3(b) 3xy2 −4x3 y''+ y' 6xyy'+3y2 −12x2 B1 B1 Differentiates 3xy2 −4x3 y'. ( ) +3y2y'−12 x2y'+2xy =0 B1 Differentiates y3−12x2y. 3xy2y''−4x3y''+6xy ( y' )2 +6y2y'−24x2y'−24xy=0 ( )( ) ( )( ) 3xy2 −4x3 12x2y'+24xy−3y2y' − 12x2y− y3 6xyy'+3y2 −12x2 y''= ( 3xy2 −4x3 )2 y''+11(−75)+3(11)−12(9)=0 M1 Substitutes (–1, 1) y''=900 A1 5 © UCLES 2021 Page 7 of 13

5(a) d B1 Differentiates (x1)y correctly. (x1)y(x1)y' y dx d  x y13 3x y12y'1 B1 Differentiates x y13 correctly. dx (1)y'3y'10 y'3 B1 Substitutes (0,0), AG. 4 3 5(b) x1y''2y' B1 Differentiates (x1)y' y. 3x y12 y''61 y'2x y10 B1 B1 Differentiates 3(x y1)21 y'. Expect 6x y11 y6x y11 yy instead of 61 y2x y1. y''233y''612 0 M1 Substitutes (0,0). 4 4 y'' 9 A1 32 y''(0) M1 A1 Finds Maclaurin’s series. y  y(0) y'(0)x x2 3 x 9 x2 2! 4 64 7 © UCLES 2022 Page 9 of 13

5(a) d B1 Differentiates (x1)y correctly. (x1)y(x1)y' y dx d  x y13 3x y12y'1 B1 Differentiates x y13 correctly. dx (1)y'3y'10 y'3 B1 Substitutes (0,0), AG. 4 3 5(b) x1y''2y' B1 Differentiates (x1)y' y. 3x y12 y''61 y'2x y10 B1 B1 Differentiates 3(x y1)21 y'. Expect 6x y11 y6x y11 yy instead of 61 y2x y1. y''233y''612 0 M1 Substitutes (0,0). 4 4 y'' 9 A1 32 y''(0) M1 A1 Finds Maclaurin’s series. y  y(0) y'(0)x x2 3 x 9 x2 2! 4 64 7 © UCLES 2022 Page 9 of 13

4(a) dy 1 B1 1 dt 1t2 dx 1 B1 1 dt 1t2 dy dy dt 1t2 1 1t2 M1 Uses chain rule.     dx dt dx 1t2 t2  1t2 1  1t2  1t2 A1 AG  1t2    t2  t2   4 © UCLES 2022 Page 8 of 15 4(b) t2  2tt  1t21 2  2t  1t2  1t2  M1 A1 Applies product or quotient rule. d dy      dtdx t4 t2 1t21 2 2  1 1t2  M1 A1 Uses chain rule. Simplification not required for A1. d2y    1t2 dx2 t5   t2 22 t2 1 1t2  t5 80 when t 3. A1 3 4 5 © UCLES 2022 Page 9 of 15

1 ex B1 Finds first derivative. y'= 1+ex ex B1 Finds second derivative. y''= ( 1+ex)2 y(0)=ln2, y'(0)= 1, y''(0)= 1 B1 Evaluates at x=0. 2 4 y=y(0)+y'(0)x+ 1 y''(0)x2+ M1 Allow 2! missing. 2! ln2+1x+1x2 A1 Decimal used for ln2 scores A0. 2 8 5 Question Answer Marks Guidance

2(a) d B1 Differentiates (x+1)y correctly. ((x+1)y)=(x+1)y'+ y dx d ( y2) =2yy' B1 Differentiates y2correctly. dx (1)y'−2+2(−2)y'=0 leading toy'=−2 B1 Substitutes (0,−2), AG. 3 3 © UCLES 2022 Page 6 of 15 2(b) (x+1)y''+2y' B1 Differentiates (x+1)y'+ y. +2yy''+2(y')2 =0 B1 Differentiates 2yy'. y''+2 ( −2) +2(−2)y''+2 ( −2)2 =0 M1 Substitutes (0,−2) into their expression with y''. 3 3 y''=− 4 A1 27 Alternative method for question 2(b) −y (x+1+2y)(−y')+ y(1+2y') B1 B1 B1 for (x+1+2y)(−y') in numerator. y'= y''= x+1+2y (x+1+2y)2 B1 for y(1+2y') in numerator. (0+1+2(−2))(2)−2 ( 1+2(−2)) M1 Substitutes (0,−2) and y'=−2 into their y''= 3 3 3 (0+1+2(−2))2 expression. y''=− 4 A1 27 4 © UCLES 2022 Page 7 of 15

1 ex B1 Finds first derivative. y'= 1+ex ex B1 Finds second derivative. y''= ( 1+ex)2 y(0)=ln2, y'(0)= 1, y''(0)= 1 B1 Evaluates at x=0. 2 4 y=y(0)+y'(0)x+ 1 y''(0)x2+ M1 Allow 2! missing. 2! ln2+1x+1x2 A1 Decimal used for ln2 scores A0. 2 8 5 Question Answer Marks Guidance

8(a) 2 et et B1 secht  tanht  et et et et  2  2  et et2 4 e2t 2e2t 4  et et2 M1 A1 Expands, gets to e2t 2e2t 4 for M1, AG. 1 et et     et et2   et et2   et et2  et et2 3 8(b) dy B1 4tanh3tsech2t dt dx tanhtsech2ttanht  tanht  sech2t1  tanh3t  M1 A1 M1 sensible attempt at derivative of x. dt dy dy dt 4tanh3tsech2t M1 A1 dy dx    4sech2t Applies chain rule, must substitute their and dx dt dx tanh3t dt dt for M1, AG. 5 8(c) d2y d dy dt 8sech2ttanht sech2t M1 A1 d2y d dy     8 Finds . For M1  csech2ttanht.AEF. dx2 dtdx dx tanh3t tanh2t dx2 dtdx Accept 8cosech2t. 8 sech2t 9 8  1tanh2t   9tanh2ttanh2t  16 M1 A1 Sets equal to 9 2 , uses identity from (a) or equivalent. tanh2t 2 2 25 Accept sinh2t16 or cosh2t 25. 9 9 x,y 8 ln3,881 A1 A1 A1 for each correct coordinate. t ln3 25 5 625 6 © UCLES 2023 Page 15 of 15

8(a) 2 et et B1 secht  tanht  et et et et  2  2  et et2 4 e2t 2e2t 4  et et2 M1 A1 Expands, gets to e2t 2e2t 4 for M1, AG. 1 et et     et et2   et et2   et et2  et et2 3 8(b) dy B1 4tanh3tsech2t dt dx tanhtsech2ttanht  tanht  sech2t1  tanh3t  M1 A1 M1 sensible attempt at derivative of x. dt dy dy dt 4tanh3tsech2t M1 A1 dy dx    4sech2t Applies chain rule, must substitute their and dx dt dx tanh3t dt dt for M1, AG. 5 8(c) d2y d dy dt 8sech2ttanht sech2t M1 A1 d2y d dy     8 Finds . For M1  csech2ttanht.AEF. dx2 dtdx dx tanh3t tanh2t dx2 dtdx Accept 8cosech2t. 8 sech2t 9 8  1tanh2t   9tanh2ttanh2t  16 M1 A1 Sets equal to 9 2 , uses identity from (a) or equivalent. tanh2t 2 2 25 Accept sinh2t16 or cosh2t 25. 9 9 x,y 8 ln3,881 A1 A1 A1 for each correct coordinate. t ln3 25 5 625 6 © UCLES 2023 Page 15 of 15

1(a) f '(x)  1x21 2 B1 Correct first derivative. f ''(x)x  1x23 2 B1 Correct second derivative. f '''(x)3x2 1x25 2   1x23 2 M1 Differentiates their f ''(x) using product rule. f(0)0 f '(0)1 f ''(0)0 f '''(0)1 M1 Evaluates their derivatives at x0. Must have attempted all three derivatives. sin1x x 1x3 A1 CWO. 6 Alternative method for question 1(a) sinyx f '(x)secy (B1) Finds first derivative. f ''(x)tanysecy2 (B1) Finds second derivative. f '''(x)  2tan2 ysec2 ysec4 y  secy3sec5 y2sec3 y (M1) Differentiates tanysecy2 using product and chain rule. f(0)0 f '(0)1 f ''(0)0 f '''(0)1 (M1) Evaluates their derivatives at x0. Must have attempted all three derivatives. sin1x x 1x3 (A1) CWO. 6 5 © UCLES 2023 Page 6 of 15 1(b) 151 B1 750 1 Partial Question Answer Guidance Marks

4(a) d  4y3 12y2y' B1 Differentiates y3 correctly. dx d  x y6 6(x y)5(1 y') B1 Differentiates x y6 correctly. dx 12  32 y'611 y'0 y' 1 B1 Substitutes (4,3), AG. 17 3 © UCLES 2023 Page 8 of 15 4(b) 2y2y''4yy2 B1 Differentiates y2y. x y5 y5x y41 y2 0 B1 B1 Differentiates (x y)51 y. 17y12  1 2 5(1)4 182 017y 96 0 M1 Substitutes (4,3). 17 17 17 96 A1 y 289 5 Partial Question Answer Guidance Marks

1 EITHER Solution 1 M1 Changes ln(x+2) or ln ( x2 +5 ) ln(x+2)=ln ( 2(1+1x)) =ln2+ln(1+1x) 2 2 so that the formula given in the list of formulae (MF19) can be applied. ln(x+2)=ln2+1x−1x2+ A1 2 8 ln ( x2 +5 ) =ln ( 5 ( 1+1x2)) =ln5+1x2 + A1 5 5 ( ln2+1x−1x2 + ) + ( ln5+1x2+ ) M1 Sums power series 2 8 5 OR Solution 2 (M1 A1) Finds first derivative. f '(x)=(x+2)−1 +2x ( x2 +5 )−1 f ''(x)=−(x+2)−2 −(2x)2( x2 +5 )−2 +2 ( x2 +5 )−1 (A1) Finds second derivative. f(0)=ln10 f '(0)= 1 f ''(0)= 3 (M1) Evaluates derivatives at zero. 2 20 ln(x+2)+ln ( x2 +5 ) =ln10+1x+ 3 x2 A1 Accept ln10 written as ln2+ln5 2 40 but do not accept decimals. WWW. 5 © UCLES 2023 Page 6 of 15

2(a) dy dx B1 =et(t+1), =−t−2 dt dt dy = dy  dt =−t2et(t+1)=−et( t3+t2) M1 A1 Uses chain rule, AG. dx dt dx 3 2(b) d   dy  =−et( 3t2 +2t ) −et( t3+t2) =−et( t3+4t2 +2t ) M1 A1 Applies product rule. dtdx d2y =t3et( t2 +4t+2 ) M1 A1 Uses chain rule, multiplies by their dt dx2 for M1. dx Alternative method for question 2(b) d2y =−et  3t2 dt +2t dt  −et( t3 +t2)dt (M1 A1) Differentiates −et( t3+t2) dx2  dx dx dx implicitly. d2y =et( −4t2 −2t−t3)dt =t3et( t2 +4t+2 ) (M1 A1) Substitutes dt =−t2. dx2 dx dx 4 © UCLES 2023 Page 7 of 15

2 e1x2 e  ex2 e  1x2   B1 Changes e1x2 or e1x so that the formula given in the list of formulae (MF19) can be applied. e1x e  1x 1x2   B1 2 e  1x2   e  1x 1x2   M1 Sums power series. 2 e1x2 e1x 2eex 3ex2  A1 2 Alternative method for question 2 f'(x)2xe1x2 e1x B1 Finds first derivative. f''(x)4x2e1x2 2e1x2 e1x B1 Finds second derivative. f(0)2e f'(0)e f''(0)3e M1 Evaluates their derivatives at zero. e1x2 e1x 2eex 3ex2  A1 2 4 © Cambridge University Press & Assessment 2024 Page 7 of 14

3(a) dy t dx 1 B1 cos1t ,  dt 1t2 dt 1t2 dy dy dt M1 A1 Uses chain rule, AG.   t 1t2 cos1t dx dt dx 3 3(b) d dy 1t2 tcos1t tcos1t M1 A1 Applies product rule.  1  2 dtdx 1t2 1t2 1t2 d2y M1 A1 Uses chain rule. 2 1t2 tcos1t dx2 4 Question Answer Marks Guidance

2 e1x2 e  ex2 e  1x2   B1 Changes e1x2 or e1x so that the formula given in the list of formulae (MF19) can be applied. e1x e  1x 1x2   B1 2 e  1x2   e  1x 1x2   M1 Sums power series. 2 e1x2 e1x 2eex 3ex2  A1 2 Alternative method for question 2 f'(x)2xe1x2 e1x B1 Finds first derivative. f''(x)4x2e1x2 2e1x2 e1x B1 Finds second derivative. f(0)2e f'(0)e f''(0)3e M1 Evaluates their derivatives at zero. e1x2 e1x 2eex 3ex2  A1 2 4 © Cambridge University Press & Assessment 2024 Page 7 of 14

3(a) dy t dx 1 B1 cos1t ,  dt 1t2 dt 1t2 dy dy dt M1 A1 Uses chain rule, AG.   t 1t2 cos1t dx dt dx 3 3(b) d dy 1t2 tcos1t tcos1t M1 A1 Applies product rule.  1  2 dtdx 1t2 1t2 1t2 d2y M1 A1 Uses chain rule. 2 1t2 tcos1t dx2 4 Question Answer Marks Guidance

3(a) d  8y3 24y2y' B1 Differentiates y3 correctly. dx d B1 Differentiates xy correctly. 2xy2xy' y dx 3(2)2 2(2)y'2(1)24(1)2y'01020y'0 B1 Substitutes (2,1) into  y'1 3x2 2xy'2y24y2y'0, AG. Must see 2 intermediary step between substitution and the final answer. 3 © Cambridge University Press & Assessment 2024 Page 7 of 15 3(b) 6x2xy''2y' B1 Differentiates 3x2 2xy'. 2y'24y2y''24y'(2y)y'0 B1 B1 Differentiates 2y24y2y'. B1 for 2y'24y2y''. B1 for 48yy'2 . (Allow alternative using quotient rule.) 6(2)2(2)y''412412 y''48(1)12 020y''260 M1 Substitutes (2,1) into their expression (must 2 2 have derivatives y' and y''). y''13 A1 OE 10 5 © Cambridge University Press & Assessment 2024 Page 8 of 15

2(a) dy = dy  dt =− 1  ( −t2) = t2 M1A1 Uses chain rule, AG. dx dt dx 1−t2 1−t2 2 2(b) d dy 2t 1−t2 +t3( 1−t2)−1 2 M1A1 Applies Quotient rule.  = dtdx 1−t2 d2y = 2t 1−t2 +t3( 1−t2)−1 2 ( −t2) =−2t3( 1−t2)−1 2 −t5( 1−t2)−3 2 M1A1 Uses chain rule. dx2 1−t2 =−t3( 1−t2)−3 2 ( 2 ( 1−t2) +t2 ) =−t3( 1−t2)−3 2 ( 2−t2) 4 © Cambridge University Press & Assessment 2024 Page 6 of 14

2(a) d ( 4y2) =8yy' B1 Differentiates y2 correctly. dx d xy'+ y B1 Differentiates ln(xy) correctly. (4ln(xy))=4 =4y'y−1+4x−1 dx xy 8 (1) y'+4(2)y'+4 (1) =0y'=−1 B1 Substitutes (2,1) into 2 2 6 2 8y'y+4y'y−1+4x−1=0, AG. 3 © Cambridge University Press & Assessment 2024 Page 6 of 13 2(b) EITHER Solution 1 (B1B1 Differentiates 8y'y+4y'y−1+4x−1=0. 8(y')2 +8y''y−4(y')2 y−2+4y−1y''−4x−2 =0 B1 for 8(y')2 +8y''y and B1 for other terms. 8(−1)2 +8y''(1)−16(−1)2 +8y''−1=0 M1) Substitutes ( 2,1) . 6 2 6 2 OR Solution 2 (B1B1 B1 for x−1( 2y+ y−1)−2( 2y'−y−2y' ) and B1 y'=−x−1( 2y+ y−1)−1 for x−2( 2y+ y−1)−1 . y''=x−1( 2y+ y−1)−2( 2y'−y−2y' ) +x−2( 2y+ y−1)−1 y''=1(1+2)−2(−1)(2−4)+1(1+2)−1 M1) Substitutes ( 2,1) and y'=−1. 2 6 4 2 6 12y''−11=0 y''= 11 A1 9 108 4 Question Answer Marks Guidance

2(a) dy = dy  dt =− 1  ( −t2) = t2 M1A1 Uses chain rule, AG. dx dt dx 1−t2 1−t2 2 2(b) d dy 2t 1−t2 +t3( 1−t2)−1 2 M1A1 Applies Quotient rule.  = dtdx 1−t2 d2y = 2t 1−t2 +t3( 1−t2)−1 2 ( −t2) =−2t3( 1−t2)−1 2 −t5( 1−t2)−3 2 M1A1 Uses chain rule. dx2 1−t2 =−t3( 1−t2)−3 2 ( 2 ( 1−t2) +t2 ) =−t3( 1−t2)−3 2 ( 2−t2) 4 © Cambridge University Press & Assessment 2024 Page 6 of 14

1 f '(x)=−(x+2)−2e (x+2)−1 B1 Finds first derivative. f ''(x)=(x+2)−4e(x+2)-1 +2(x+2)−3 e(x+2)-1 = 2x+5 e (x+2)−1 B1 Finds second derivative. (x+2)4 f(0)=e 1 2 f '(0)=−1e 1 2 f ''(0)= 5 e 1 2 *M1 Evaluates derivatives at zero. Must have 4 16 attempted second derivative using the product rule. e (x+2)−1 =e 1 2 −1e 1 2x+ 5 e 1 2x2 DM1 A1 4 32 5 Question Answer Marks Guidance

3(a) d ( 9y2) =18yy' B1 Differentiates y2 correctly. dx d ( −3sinh−1(xy) ) =−3 xy'+ y M1 A1 Differentiates sinh−1(xy) correctly. dx 1+x2y2 18 y'−3 4y'+1 3 =0 y'=−1 A1 Substitutes( 4,1) and shows at least one 3 5 2 3 3 intermediate step, AG. 4 © Cambridge University Press & Assessment 2025 Page 10 of 17 3(b) 18yy''+18(y')2 B1 Differentiates 18yy'  1+x2y2 (xy''+2y')−(xy'+ y)2( 1+x2y2)−1 2(xy)  B1 B1 Differentiates −3 xy'+ y .   −3 1+x2y2  1+x2y2      5(4y''−1)−(5)2(3)(4) M1 Substitutes ( 4,1) . and y=− 1 18 y''+18−33 3 5 3 =0 3 2 3 4 25 9 45(4y''−1)−60 18 y''+18− =0 3 4 25 −6 y''+87 =0 y''= 29 A1 5 10 4 5 © Cambridge University Press & Assessment 2025 Page 11 of 17

4(a) x=0(t−1)(t2 +1)=0 B1 Factorises or considers other roots of x=0. 1 © Cambridge University Press & Assessment 2025 Page 11 of 17 4(b) dx dy B1 =3t2 −2t+1 =(t+1)et dt dt dy dy dt dy (t+1)et M1 A1 Applies chain rule, AG. =   = dx dt dx dx 3t2 −2t+1 3 4(c) d  (t+1)et  ( 3t2 −2t+1 )( (t+1)et +et) −(t+1)et(6t−2) *M1 Differentiates dy with respect to t.  = dx dt 3t2 −2t+1 ( 3t2 −2t+1 )2 d2y ( 3t2 −2t+1 ) (t+2)et −(t+1)(6t−2)et M1 A1 Applies chain rule. = dx2 ( 3t2 −2t+1 )3 y(0)=e, y'(0)=e, y''(0)=−1e A1 Evaluates derivatives at t=1. 4 y''(0) DM1 A1 Finds Maclaurin’s series. For M1, must have y= y(0)+ y'(0)x+ x2 =e+ex−1ex2 2! 8 attempted second derivative. 6 © Cambridge University Press & Assessment 2025 Page 12 of 17

2(a) dy dy dt 1+sinht M1 A1 Applies chain rule. =  = dx dt dx cosht 2 2(b) d d t    1+ co s s i h nh t t   = cosht(cosht c ) o − sh (1 2 + t sinht)sinht = 1 c − o s s i h n 2 h t t M1 A1 Finds d d t    d d y x    . d2y d dy dt 1−sinht M1 A1 Applies chain rule, AG. =    = dx2 dtdx dx cosh3t 4 2(c) x2 M1 A1 Finds Maclaurin’s series. M1 for y(0), y'(0) and y= y(0)+xy'(0)+ y''(0)=1+x+ 1x2 2! 2 y''(0) evaluated. A0 if 2! not simplified to 2. 2 © Cambridge University Press & Assessment 2025 Page 9 of 19

3(a) d ( x2y ) = x2y'+2xy B1 Differentiates x2y correctly. dx d ( (x+ y)3) =3(x+ y)2(1+ y') M1 Differentiates (x+ y)3 correctly. dx y'−4+3(1+ y')=0 y'= 1 A1 Some evidence of substituting (−1,2), into an 4 equation set to zero. AG. 3 3(b) x2y''+2xy'+2xy'+2y B1 B1 Differentiates x2y'+2xy. +3(x+ y)2 y''+6(1+ y')2(x+ y)=0 B1 B1 Differentiates 3(x+y)2(1+y'). (−1)2y''+4(−1)(1)+2(2)+3y''+6(5)2=0 M1 Substitutes (−1,2) and y'= 1 into their 4 4 4 expression with at least a y''in their expression y''=−99 A1 32 6 © Cambridge University Press & Assessment 2025 Page 10 of 17

2(a) dy dy dt 1+sinht M1 A1 Applies chain rule. =  = dx dt dx cosht 2 2(b) d d t    1+ co s s i h nh t t   = cosht(cosht c ) o − sh (1 2 + t sinht)sinht = 1 c − o s s i h n 2 h t t M1 A1 Finds d d t    d d y x    . d2y d dy dt 1−sinht M1 A1 Applies chain rule, AG. =    = dx2 dtdx dx cosh3t 4 2(c) x2 M1 A1 Finds Maclaurin’s series. M1 for y(0), y'(0) and y= y(0)+xy'(0)+ y''(0)=1+x+ 1x2 2! 2 y''(0) evaluated. A0 if 2! not simplified to 2. 2 © Cambridge University Press & Assessment 2025 Page 9 of 19

7(a)(i) sec2 1t B1 Differentiates x with respect to t. AG. x=4 2 +3cosec2t−3 tan1t 2 x=8cosect+3cosec2t−3=8cosect+3cot2t 1 7(a)(ii) y=6cosect−4cot2t B1 Differentiates y with respect to t. 1 7(a)(iii) x2+ y2 = ( 8cosect+3cot2t )2 + ( 6cosect−4cot2t )2 M1 A1 Expands and simplifies x2+ y2. A1 for reaching 100cosec2t+25cot4t . =100cosec2t+25cot4t =100cosec2t+25 ( cosec2t−1 )2 =25cosec4t+50cosec2t+25 =25 ( cosec2t+1 )2 *M1 A1 Factorises x2+ y2. ALT: 25 ( cot2t+2 )2 . 5 1 3 π cosec2t+1dt =5−cott+t1 3 π =5 ( − 1 +1π+ 3−1π ) = 5π+10 3 DM1 A1 Applies correct formula for arc length and 1 6 π 1 6 π 3 3 6 6 3 integrates. Answer in terms of π and 3 for A1. 6 © Cambridge University Press & Assessment 2025 Page 16 of 19 7(b) y 6cosect−4cot2t 6sint−4cos2t 6sect−4cott B1 Finds first derivative. = = = x 8cosect+3cot2t 8sint+3cos2t 8sect+3cott d 6cosect−4cot2t M1 dy Differentiates with respect to t.   dt8cosect+3cot2t  dx Enough terms for M1. ( 8csct+3cot2t )( −6cottcsct+8cottcsc2t ) + ( 6csct−4cot2t )( 6csc2tcott+8cottcsct ) = ( 8cosect+3cot2t )2 or d 6sint−4cos2t   dt8sint+3cos2t  ( 8sint+3cos2t )(6cost+8costsint)− ( 6sint−4cos2t )(8cost−6costsint) = ( 8sint+3cos2t )2 100costsin2t+50cos3t 50cost ( sin2t+1 ) 50cottcosect ( cosec2t+1 ) A1 = = ( 8sint+3cos2t )2 ( 8sint+3cos2t )2 ( 8cosect+3cot2t )2 d2y d 6sint−4cos2t dt 50sin2tcost ( sin2t+1 ) M1 A1 Applies chain rule. =   = dx2 dt8sint+3cos2t  dx ( 8sint+3cos2t )3 50cottcosect ( cosec2t+1 ) = ( 8cosect+3cot2t )3 5 © Cambridge University Press & Assessment 2025 Page 17 of 19

4(a) d d x ( ( x − y ) 5 ) = 5 ( x − y ) 4 ( 1 − y )' B1 Differentiates ( x − y ) 5 correctly. d d x ( ln ( x y ) ) = x y '+ x y y = y ' y − 1 + x − 1 B1 Differentiates ln(xy) correctly. 5 ( 1 − y )' − 12 y '− 1 = 0  y ' = 81 1 B1 Substitutes ( − 1 , − 2 ) into 5 ( x − y ) 4 ( 1 − y )' + y ' y − 1 + x − 1 = 0 , AG. CWO. Might see y (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 . 3 Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) EITHER: © Cambridge University Press & Assessment 2026 Page 12 of 18 − 5 ( x − y ) 4 y ''+ 2 0 ( 1 − y )' 2 ( x − y ) 3 (B1 B1 Differentiates 5 ( x − y ) 4 ( 1 − y )' . − ( y ') 2 y − 2 + y − 1 y ''− x − 2 = 0 B1 Differentiates y ' y − 1 + x − 1 . Might see x y '+ x y y differentiated using product or quotient rules. xy(xy''+2y')−(xy'+ y)2 E.g. . (xy)2 Or − ( x y ) − 2 ( x y '+ y ) 2 + ( x y ) − 1 ( x y ''+ 2 y )' . − 5 (1 ) 4 y ''+ 2 0 ( 1 − 81 1 ) 2 ( 1 ) 3 − 14 ( 81 1 ) 2 − 12 y ''− 1 = 0 M1 Substitutes ( − 1 , − 2 ) and y'= 8 . 11 − 1 12 y ''+ 4 3 1 2 1 = 0  y '' = 1 83 63 1 A1) OR: Alternative method for question 4(b) y (' − y − 2 y '− 2 0 ( x − y ) 3 ( 1 − y )' ) + y ( '' y − 1 − 5 ( x − y ) 4 ) (B1 B1 Differentiates both sides of y (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 . = − 2 0 ( x − y ) 3 ( 1 − y )' + x − 2 B1 ( 81 1 ) ( − 14  81 1 − 2 0  31 1 ) + y ( '' − 12 − 5 ) = − 2 0  31 1 + 1 M1 Substitutes ( − 1 , − 2 ) and y'= 8 . 11 − 1 12 y ''+ 4 3 1 2 1 = 0  y '' = 1 83 63 1 A1) Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) OR: Second alternative method for question 4(b) ( 5(x−y)4 −y−1 )( 20(x−y)3(1−y')−x−2 ) (B1 B1 Numerator of derivative of 5(x− y)4 +x−1 − ( 5(x−y)4 +x−1 )( 20(x− y)3(1− y')+ y−2y' ) y'= 5(x− y)4 − y−1 . © Cambridge University Press & Assessment 2026 Page 13 of 18 ( 5 ( x − y ) 4 − y − 1 ) 2 B1 Denominator of derivative of y ' = 5 5 ( ( x x − − y y ) ) 4 4 + − x y − − 1 1 . Can be implied by ( 1 12 ) 2 . y '' = ( 5 + 12 ) ( 2 0  31 1 − 1 ( 5 ) − + ( 12 5 ) − 2 1 ) ( 2 0  31 1 + 21 1 ) M1 Substitutes ( − 1 , − 2 ) and y'= 8 into their 11 expression. y '' = 1 83 63 1 A1) Available marks 5 Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) d d x ( ( x − y ) 5 ) = 5 ( x − y ) 4 ( 1 − y )' B1 Differentiates ( x − y ) 5 correctly. d d x ( ln ( x y ) ) = x y '+ x y y = y ' y − 1 + x − 1 B1 Differentiates ln(xy) correctly. 5 ( 1 − y )' − 12 y '− 1 = 0  y ' = 81 1 B1 Substitutes ( − 1 , − 2 ) into 5 ( x − y ) 4 ( 1 − y )' + y ' y − 1 + x − 1 = 0 , AG. CWO. Might see y (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 . 3 Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) EITHER: © Cambridge University Press & Assessment 2026 Page 12 of 18 − 5 ( x − y ) 4 y ''+ 2 0 ( 1 − y )' 2 ( x − y ) 3 (B1 B1 Differentiates 5 ( x − y ) 4 ( 1 − y )' . − ( y ') 2 y − 2 + y − 1 y ''− x − 2 = 0 B1 Differentiates y ' y − 1 + x − 1 . Might see x y '+ x y y differentiated using product or quotient rules. xy(xy''+2y')−(xy'+ y)2 E.g. . (xy)2 Or − ( x y ) − 2 ( x y '+ y ) 2 + ( x y ) − 1 ( x y ''+ 2 y )' . − 5 (1 ) 4 y ''+ 2 0 ( 1 − 81 1 ) 2 ( 1 ) 3 − 14 ( 81 1 ) 2 − 12 y ''− 1 = 0 M1 Substitutes ( − 1 , − 2 ) and y'= 8 . 11 − 1 12 y ''+ 4 3 1 2 1 = 0  y '' = 1 83 63 1 A1) OR: Alternative method for question 4(b) y (' − y − 2 y '− 2 0 ( x − y ) 3 ( 1 − y )' ) + y ( '' y − 1 − 5 ( x − y ) 4 ) (B1 B1 Differentiates both sides of y (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 . = − 2 0 ( x − y ) 3 ( 1 − y )' + x − 2 B1 ( 81 1 ) ( − 14  81 1 − 2 0  31 1 ) + y ( '' − 12 − 5 ) = − 2 0  31 1 + 1 M1 Substitutes ( − 1 , − 2 ) and y'= 8 . 11 − 1 12 y ''+ 4 3 1 2 1 = 0  y '' = 1 83 63 1 A1) Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) OR: Second alternative method for question 4(b) ( 5(x−y)4 −y−1 )( 20(x−y)3(1−y')−x−2 ) (B1 B1 Numerator of derivative of 5(x− y)4 +x−1 − ( 5(x−y)4 +x−1 )( 20(x− y)3(1− y')+ y−2y' ) y'= 5(x− y)4 − y−1 . © Cambridge University Press & Assessment 2026 Page 13 of 18 ( 5 ( x − y ) 4 − y − 1 ) 2 B1 Denominator of derivative of y ' = 5 5 ( ( x x − − y y ) ) 4 4 + − x y − − 1 1 . Can be implied by ( 1 12 ) 2 . y '' = ( 5 + 12 ) ( 2 0  31 1 − 1 ( 5 ) − + ( 12 5 ) − 2 1 ) ( 2 0  31 1 + 21 1 ) M1 Substitutes ( − 1 , − 2 ) and y'= 8 into their 11 expression. y '' = 1 83 63 1 A1) Available marks 5 Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/