Let `x^cosy + y^cosx = 5`, then

Options

A. at ` x = 0 , y = 0, y'=0`
B. at ` x = 0 , y= 1 , y'= 0 `
C. at ` x = y = 1, y'= - 1`
D. at ` x = 1 , y=0 , y'=1`

Answer

C

Explanation

`x^(cos y) + y^(cos x) = 5 `
` rArr e^(cos y log_(e)x+ )e^(cos x loge y) = 5`
` therefore e^(cos y loge x){(cos y)/(x) - log_(e) x sin y (dy)/(dx)} + e^(cos x log y){(cos x )/(y) (dy)/(dx) - sin x log_(e) y} = 0 `
Put x = y = 1 ,
then ` (cos 1 - 0) + 1 xx (cos 1(dy)/(dx) - 0) = 0 `
` therefore (dy)/(dx) = - 1`
or ` y'= - 1`