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

2024
Views:

46

Pages:

2
Author:

Shahbaz A.

IMPLICIT DIFFERENTIATION 14-07-2024 shahbaz ahmed July 2024 Ex 3.7 THOMAS CALCULUS Twelfth Edition Use implicit differentiation to find Q5 dy dx x2 (x − y)2 = x2 − y 2 Solution d(x2 (x−y)2 ) dx = d(x2 −y 2 ) dx 2 2 ) + (x − y)2 d(x x2 d(x−y) dx dx = d(x2 ) dx − d(y 2 ) dx (x2 )(2)(x − y) d(x−y) + 2x(x − y)2 = 2x − dx (x2 )(2)(x − y)[ dx dx − (x2 )(2)(x − y)[1 − 2x2 (x − y)(1 − dy dx ) (2x3 − 2x2 y)(1 − dy dx ] dy dx ] d(y 2 ) dy dy dx + 2x(x − y)2 = 2x − d(y 2 ) dy dy dx dy + 2x(x − y)2 = 2x − 2y dx dy + 2x(x − y)2 = 2x − 2y dx dy dx ) dy + 2x(x − y)2 = 2x − 2y dx dy dy dy 2x3 − 2x3 dx − 2x2 y + 2x2 y dx + 2x(x − y)2 − 2x + 2y dx =0 dy [−2x3 + 2x2 y + 2y] dx = −2x3 + 2x2 y − 2x(x − y)2 + 2x dy [−2x3 + 2x2 y + 2y] dx = −2x3 + 2x2 y − 2x(x2 − 2xy + y 2 ) + 2x dy 2[−x3 + x2 y + y] dx = −2x3 + 2x2 y − 2x3 + 4x2 y − 2xy 2 + 2x dy 2[−x3 + x2 y + y] dx = −4x3 + 6x2 y − 2xy 2 + 2x 1 dy 2[−x3 + x2 y + y] dx = 2(−2x3 + 3x2 y − xy 2 + x) dy [−x3 + x2 y + y] dx = −2x3 + 3x2 y − xy 2 + x dy = −2x3 + 3x2 y − xy 2 + x [x2 y − x3 + y] dx dy dx = −2x3 +3x2 y−xy 2 +x x2 y−x3 +y 2