If `log_(10)((x^(3)-y^(3))/(x^(3)+y^(3)))=2`, then `(dy)/(dx)=`

Options

A. `(x)/(y)`
B. `-(y)/(x)`
C. `-(x)/(y)`
D. `(y)/(x)`

Answer

D

Explanation

Given,
`log_(10)((x^(3)-y^(3))/(x^(3)+y^(3)))=2`
`implies(x^(3)-y^(3))/(x^(3)+y^(3))=10^(2)=100`
`impliesx^(3)-y^(3)=100(x^(3)+y^(3))`
`implies101y^(3)=-99x^(3)`
On differentiating both sides w.r.t. x, we get
`101xx3y^(2)(dy)/(dx)=-99*(3x^(2))implies101y^(2)(dy)/(dx)=-99x^(2)`
On multipying by x both sides, we get
`implies101xy^(2)(dy)/(dx)=-99x^(3)`
`implies(dy)/(dx)=(-99x^(3))/(101xy^(2))`
`implies(dy)/(dx)=(101y^(3))/(101xy^(2))" "[because-99x^(3)=101y^(3)]`
`implies(dy)/(dx)=(y)/(x)`