If `x^(2) y^(5) = (x + y)^(7) , " then " (d^(2)y)/(dx^(2))` is equal to

Options

A. `y //x ^(2)`
B. `x//y`
C. 1
D. 0

Answer

D

Explanation

Given , `x^(2) y^(2) = (x + y)^(7)`
Taking log on both sides , we get
2 log x + 5 log y = 7 log ( x + y)
On differentiating , we get
`(2)/(x) + (5)/(y) (dy)/(dx) = (7 )/( x + y) = ( 1 + (dy)/(dx))`
`implies (dy)/(dx) ((7)/(x+y) - (5)/(y)) = (2)/(x) - (7)/(x + y)`
`implies (dy)/(dx) = (y)/(x) " " .... (i)`
Again , differentiating , we get
`(d^(2) y)/(dx^(2)) = ( x ""(dy)/(dx) - y)/(x^(2))`
` = (x * (y //x) - y)/(x^(2)) `[ from Eq. (i) ]
= 0