if `2^x+2^y=2^(x+y)` then the value of `(dy)/(dx)` at `x=y=1`

Options

A. 0
B. -1
C. 1
D. 2

Answer

B

Explanation

Given that `2^(x)+2^(y)=2^(x+y)`
On differentiating w.r.t `x` we get
`2^(x)(log2)+2^(y)(log2)(dy)/(dx)`
`=2^(x+y)(log2)(1+(dy)/(dx))`
`implies2^(x)+2^(y) (dy)/(dx)`
`=2^(x)+2^(y)(dy)/(dx)=2^(x+y)+2^(x+y)((dy)/(dx))`
`implies(dy)/(dx)(2^(y)-2^(x+y))=2^(x+y)-2^(x)`
`implies (dy)/(dx)=(2^(x+y)-2^(x))/(2^(y)-2^(x+y))`
`:.((dy)/(dx))_(x=y=1)=(2^(2)-2)/(2-2^(2))=2/(-2)=-1`