University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

480

Pages:

4
Author:

Shahbaz A.

Q 23 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 153 shahbaz ahmed August 2024 Q23 Use implicit differentiation to find √ 2 y =x−y Solution √ 2 y =x−y d √ d [2 y] = [x − y] dx dx d √ dy dx dy 2 [ y] = − dy dx dx dx 2 d 1 dy dy [y 2 ] =1− dy dx dx 1 1 dy dy 2( )[y 2 −1 ] =1− 2 dx dx 1 dy dy [y − 2 ] =1− dx dx 1 dy dx and then d2 y dx2 1 dy dy =1− √ y dx dx 1 dy dy + =1 √ y dx dx 1 dy [ √ + 1] =1 y dx √ 1 + y dy [ √ ] =1 y dx √ y dy = √ dx 1 + y dy dx = √ y √ 1+ y √ y dy = √ dx 1 + y (1 + √ dy √ y) = y dx d d √ √ dy [(1 + y) ] = [ y] dx dx dx dy d d √ dy √ d dy √ [1 + y] = [ y] (1 + y) [ ] + dx dx dx dx dy dx 1 dy d 1 dy √ d2 y dy d (1 + y) 2 + [1 + y 2 ] = [y 2 ] dx dx dy dx dy dx 2 dy d d 1 1 1 dy √ d2 y (1 + y) 2 + ( )2 [ (1) + (y 2 )] = y 2 −1 dx dx dy dy 2 dx dy 1 1 1 1 dy √ d2 y (1 + y) 2 + ( )2 [0 + (y 2 −1 )] = y − 2 dx dx 2 2 dx 1 dy 1 1 1 dy √ d2 y (1 + y) 2 + ( )2 [ (y − 2 )] = y − 2 dx dx 2 2 dx dy 1 1 dy √ d2 y (1 + y) 2 + ( )2 [ √ ] = √ dx dx 2 y 2 y dx 1 dy dy 1 √ d2 y (1 + y) 2 = √ − ( )2 √ dx 2 y dx dx 2 y Putting √ y dy = √ dx 1 + y √ √ y y 2 1 1 √ d2 y (1 + y) 2 = √ [ √ ]−( √ ) √ dx 2 y 1+ y 1+ y 2 y √ ( y)2 1 1 1 √ d2 y (1 + y) 2 = [ √ ]− √ 2 √ dx 2 1+ y (1 + y) 2 y √ 1 1 1 ( y) √ d2 y (1 + y) 2 = [ √ ]− √ dx 2 1+ y 2 (1 + y)2 3 √ y 1 1 √ d2 y (1 + y) 2 = √ [1 − √ ] dx 21 + y (1 + y) √ √ 1 1 1+ y− y √ d2 y (1 + y) 2 = √ √ dx 2 1+ y 1+ y 1 1 √ d2 y (1 + y) 2 = √ dx 2 (1 + y)2 1 1 d2 y = √ dx2 2 (1 + y)3 4