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

2023
Views:

29

Pages:

4
Author:

Shahbaz A.

Q 20 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 153 shahbaz ahmed August 2024 Q20 Use implicit differentiation to find 2 2 2 2 x3 + y 3 = 1 Solution x3 + y 3 = 1 2 d d 2 [x 3 + y 3 ] = [1] dx dx d 2 d 2 x3 + y 3 = 0 dx dx 2 2 −1 d 2 dy x3 + y 3 =0 3 dy dx 1 dy dx and then d2 y dx2 2 − 1 2 2 −1 dy x 3 + y3 =0 3 3 dx 2 1 2 − 1 dy y 3 = − x− 3 3 dx 3 1 y− 3 1 dy = −x− 3 dx 1 dy y3 =− 1 dx x3 Or 1 dy = −y 3 dx d 1 dy d 1 [x 3 ] = − [y 3 ] dx dx dx 1 x3 1 x3 d dy dy d 1 d 1 dy [ ]+ [x 3 ] = − [y 3 ] dx dx dx dx dy dx 1 x3 d2 y dy 1 1 −1 1 1 −1 dy 3 + ( )x = −[ y3 ] dx2 dx 3 3 dx 2 d2 y dy 1 − 2 1 2 dy 3 = −[ y − 3 ] x + ( )x dx2 dx 3 3 dx 1 3 1 − 2 dy dy 1 − 2 d2 y = − y 3 − ( )x 3 x dx2 3 dx dx 3 1 3 2 d2 y 1 dy − 2 3 + x− 3 ] x = − [y dx2 3 dx 1 3 1 dy 1 1 d2 y = − + ] x [ 2 dx2 3 dx y 3 x 32 1 3 2 2 d2 y 1 dy x 3 + y 3 x = − [ ] dx2 3 dx x 23 y 32 1 3 2 2 Since x 3 + y 3 = 1 d2 y 1 dy 1 x =− [ ] 2 dx 3 dx x 23 y 32 1 3 d2 y 1 dy 1 = − [ ] 2 dx2 3 dx x 3 x 13 y 32 d2 y 1 dy 1 = − [ ] dx2 3 dx xy 23 Putting 3 1 dy y3 =− 1 dx x3 1 1 y3 1 d2 y = − × − 1 [ 2 ] dx2 3 x 3 xy 3 d2 y 1 1 = dx2 3 x 43 y 13 4