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

2024
Views:

422

Pages:

3
Author:

Shahbaz A.

differentiating implicitly shahbaz ahmed July 2024 Ex 3.7 THOMAS calculus Twelfth Edition Use implicit differentiation to find Q1 x2 y + xy 2 = 6 d(x2 y+xy 2 ) dx d(x2 y) dx + = d dx (6) d(xy 2 ) dx =0 2 2 ) d(y ) dy 2 dx x2 dx + y d(x dx + x dx + y dx = 0 2 ) dy dy 2 x2 dx + 2xy + x d(y dy dx + y = 0 dy dy x2 dx + 2xy + 2xy dx + y2 = 0 dy dy x2 dx + 2xy dx = −2xy − y 2 dy (x2 + 2xy) dx = −2xy − y 2 dy x(x + 2y) dx = −y(2x + y) 1 dy dx dy dx y(2x+y) = − x(x+2y) Q2 x3 + y 3 = 18xy d(x3 +y 3 ) dx = d(x3 ) dx d(y 3 ) dx + 3x2 + d(18xy) dx = 18 d(xy) dx d(y 3 ) dy dy dx dy dx = 18[x dx + y dx ] dy dy 3x2 + 3y 2 dx = 18[x dx + y] dy dy − 18x dx = 18y − 3x2 3y 2 dx dy (3y 2 − 18x) dx = 18y − 3x2 dy 3(y 2 − 6x) dx = 3(6y − x2 ) dy = (6y − x2 ) (y 2 − 6x) dx dy dx = 6y−x2 y 2 −6x Q3 2xy + y 2 = x + y d(2xy+y 2 ) dx = d(2xy) dx + d(y 2 ) dx = 2 d(xy) dx + d(y 2 ) dx =1+ d(x+y) dx dy 2(x dx + y dx dx ) + dx dx + dy dx dy dx d(y 2 ) dy dy dx =1+ dy dy 2x dx + 2y + 2y dx =1+ dy dx dy dx 2 dy dy + 2y dx − 2x dx dy dx = 1 − 2y dy (2x + 2y − 1) dx = 1 − 2y dy dx = 1−2y 2x+2y−1 3