`"If "y= xlog ((x)/(a+bx))," then "x^(3)(d^(2)y)/(dx^(2))=`

Options

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

Explanation

`"From the given relation, "(y)/(x)=log x-log (a+bx)`
Differentiating w.r.t. x, we get
`((x(dy)/(dx))-y)/(x^(2))=(1)/(x)-(b)/(a+bx)=(a)/(x(a+bx))`
`therefore" "x(dy)/(dx)-y=(ax)/(a+bx)" (1)"`
Differentiating again w.r.t. x, we get
`x(d^(2)y)/(dx^(2))+(dy)/(dx)-(dy)/(dx)=((a+bx)a-ax.b)/((a+bx)^(2))`
`"or "x(d^(2)y)/(dx^(2))=(a^(2))/((a+bx)^(2))`
`"or "x^(3)(d^(2)y)/(dx^(2))=(a^(2)x^(2))/((a+_bx)^(2))=(x(dy)/(dx)-y)^(2)" [by (1)]"`