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

2023
Views:

320

Pages:

3
Author:

Shahbaz A.

Q 24 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 153 shahbaz ahmed August 2024 1 Q24 Use implicit differentiation to find dy dx y 3 + 3x2 + x3 = 0 Solution y 3 + 3x2 + x3 = 0 d 3 d [y + 3x2 + x3 ] = [0] dx dx d 3 d d [y + [3x2 ] + [x3 ] = 0 dx dx dx d d d 3 dy [y ] + [3x2 ] + [x3 ] = 0 dy dx dx dx dy + 6x + 3x2 = 0 dx dy 3y 2 = −3x2 − 6x dx 3y 2 1 and then d2 y dx2 dy = −3x[x + 2] dx dy = −x[x + 2] y2 dx 3y 2 dy x(x + 2) =− dx y2 dy dx = − x(x+2) y2 dy x(x + 2) =− dx y2 y2 dy = −x(x + 2) dx d 2 dy d [y ] = − [x2 + 2x] dx dx dx d dy dy d 2 d d y2 [ ] + (y ) = −[ (x2 ) + (2x)] dx dx dx dx dx dx y2 dx d2 y dy d 2 dy + (y ) = −[2x + 2 ] dx2 dx dy dx dx 2 2d y y dx2 y2 +( dy 2 d 2 ) (y ) = −2[x + 1] dx dy d2 y dy 2 + ( ) (2y) = −2[x + 1] dx2 dx 2 Putting dy x(x + 2) =− dx y2 2 2d y y dx2 + (− 2 2d y y dx2 y2 x(x + 2) 2 ) (2y) = −2[x + 1] y2 x2 (x + 2)2 + 2( ) = −2[x + 1] y3 2x2 (x + 2)2 d2 y = −2(x + 1) − dx2 y3 2(x + 1) 2x2 (x + 2)2 d2 y =− − dx2 y2 y5 3