2022 Chp 5 Techniques of Differentiation BMQ & APQ Sol (RVHS)

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 5.1 Differentiation Techniques and Graphical Analysis (Solutions to Tutorial and Assignment) 1 Solutions to Tutorial 5A: Differentiation Techniques and Graphical Analysis Basic Mastery Questions

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 5.1 Differentiation Techniques and Graphical Analysis (Solutions to Tutorial and Assignment) 13 Additional Practice Questions 1a.        1 1 2 2 3 9sin cos 3 cos cos 3 . 1 9 1 9 d x x xdx x x        1b. d d 1 cos , sind d x y      d d d sin d d d 1 cos y y x x        At point where ,  d 1 d 2 y x  sin 1 1 cos 2 2sin 1 cos 2sin cos 1 (shown)               2(a) 2 3e 4yxy x    2 2 2 d d2 3e 4d d d d 42 3e 4d d 2 3e y y y y yy xy x x y y yxy yx x xy          (b) 2 2 2 d d d 1 2 d d2 2d v ( 1) (1 ) u vv ux t t t t t t t                  d cos cotd sin y t tt t  2 2d d d (1 ) 1 cot (1 ) cotd d d 2 2 y y t t t t tx t x         (c)     1 2 2 2 1 1 2 22 2 2 2 d 1 2 cos1 1d 1 2 1 cos cos 1 11 1 1 (shown)1 y x x xx x x x y xx x x x x xx x x x x y x                            

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 5.1 Differentiation Techniques and Graphical Analysis (Solutions to Tutorial and Assignment) 14 3. (a) t tx  1 )ln(costy 22 )1( 1 )1( )1( tt tt dt dx     tt t dt dy tancos sin  ttdx dt dt dy dx dy tan)1( 2 (b) xxey y 3sin 1  3 1 1 2   yy edx dyxedx dy y   22 131 yedx dyxeydx dy yy  2 2 11 1)3( yxe ye dx dy y y  

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 5.1 Differentiation Techniques and Graphical Analysis (Solutions to Tutorial and Assignment) 15 (c) Shape + Minimum Point x-intercept at -2: Asymptotes: x = 2, y = 0 4 )ln(22 xyxy  Differentiate implicitly w.r.t x:        x y yxxx y d d1122d d = x y yxx d d222  When x = 1, y = 1, x y x y d d222d d   4d d x y Differentiate implicitly w.r.t. x: x y yx y x y yxx y d d2 d d d d222d d 22 2 22 2  )4)(2)(4(d d222d d 2 2 2 2  x y x y  32d d 2 2 x y –2 x = 2 y x 0 y = f’ (x)