If y = In `((x)/(a + bx))^(x), "then " x^(3)(d^(y))/(dx^(2))` is equal to

Options

A. `((dy)/(dx) + x)^(2)`
B. `((dy)/(dx) - y)^(2)`
C. `(x(dy)/(dx) + y)^(2)`
D. `(x(dy)/(dx) - y)^(2)`

Answer

D

Explanation

`because y = "In" ((x)/(a + bx ))^(x) = x ("In x - In"(a + bx))`
or ` ((y)/(x)) = "In x - In(a + bx)"`
On differentiating both sides w.r.t.x , we get
`((x(dy)/(dx) - y.1)/(x^(2))) = (1)/(x)- (b)/(a + bx) = (a)/(x (a + bx)) " `"...(i)
or ` (x(dy)/(dx) - y)= ((ax)/(a + bx))`
On taking long both sides , we get
In ` (x(dy)/(dx) - y) = "In (ax) - In (a + bx)"`
On differentiating both sides w.r.t.x, we get
`(x(d^(2)y)/(dx^(2) ) + (dy)/(dx) - (dy)/(dx))/((x(dy)/(dx) - y))=(1)/(x) - (b)/(a + bx)`
` = (a)/(x (a + bx)) = (x(dy)/(dx)-y)/(x^(2)) `[ from Eq. (i) ]
or `x^(3) (d^(2) y)/(dx^(2) ) = (x(dy)/(dx)-y)^(2)`