If `y = ((ax + b)/(cx + d)), "then" 2 (dy)/(dx).(d^(3) y)/(dx^(3)) ` is equal to

Options

A. `((d^(2) y)/(dx^(2)))^(2) `
B. `3(d^(2) y)/(dx^(2))`
C. `3((d^(2) y)/(dx^(2)))^(2)`
D. ` 3 (d^(2) x)/(dy^(2))`

Answer

C

Explanation

` therefore y = ((ax + b)/(cx + d)) or cxy + dy = ax + b `
On differentiating both sides w.r.t.x, we get
`c{x (dy)/(dx) + y .1} + d (dy)/(dx) = a `
` rArr x (dy)/(dx) + y + ((d)/(c)) (dy)/(dx) = ((a)/(c))`
Again, differentiating both sides w.r.t.x, we get
`x (d^(2) y)/(dx^(2)) +(dy)/(dx) + (dy)/(dx) + ((d)/(c))(d^(2)y)/(dx^(2)) = 0`
or ` x + (2(dy)/(dx))/(((d^(2) y)/(dx^(2)))) + d/c= 0 `
Again, differentiating both sides w.r.t.x, we get
`1 + ((d^(2) y)/(dx^(2)).2(d^(2)y)/(dx^(2)) - 2(dy)/(dx) .(d^(3) y)/(dx^(3)))/(((d^(2) y)/(dx^(2)))^(2))+ 0 = 0 `
` rArr 2 (dy)/(dx).(d^(3)y)/(dx^(3)) = 3 ((d^(2) y)/(dx^(2)))^(2)`