If `2^x+2^y=2^(x+y),` then `(dy)/(dx)` is equal to

Options

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

Answer

C

Explanation

On differentiating both sides w.r.t.x , we get
`2^(x) log 2 + 2^(y) log 2(dy)/(dx) = 2^(x +y) log2(1 + (dy)/(dx))`
`rArr 2^(x) + 2^(y) (dy)/(dx) = 2^(x +y) (1 + (dy)/(dx))`
`rArr 2^(x) - 2^(y) (dy)/(dx) = 2^(x + y) (1 + (dy)/(dx))`
`rArr 2^(x) - 2^(x +y) = (dy)/(dx) (2^(x + y) - 2^(y))`
` rArr 2^(x -y) ((1 -2^(2)))/((2^(x) -1)) = (dy)/(dx) `
`rArr 2^(x-y) ((2^(y) -1)/(1-2^(x))) = (dy)/(dx)`