If `y^x=e^(y-x)` , prove that `(dy)/(dx)=((1+logy)^2)/(logy)`

Answer

N/a

Explanation

We have, `y^(x) = e^(y-x)`
`rArr logy^(x)=loge^(y-x)`
`rArr xlogy = y-x.log_(e)=(y-x), [:' log_(e) = 1]`
Now, differentiating w.r.t.x, we get
` rArr d/dylogy.(dy)/(dx) = (d)/(dx) ((y-a))/(x)`
`rArr 1.(ty).(dy)/(dx) = (x.(d)/(dx)(y-x)-(y-x).(d)/(dx).x)/(x^(2))`
`rArr (1)/(y)(dy)/(dx) = (x((dy)/(dx) - 1)-(y-x))/(x^(2))`
`rArr (x^(2))/(y).(dy)/(dx)=x'(dy)/(dx)-x-y+x`
`rArr (dy)/(dx)((x^(2))/(y)-x)= - y`
`:. (dy)/(dx) = (-y^(2))/(x^(2)-xy) = (-y^(2))/(x(x-y))`
`= (y^(2))/(x(y-x)).(x)/(x)= (y^(2))/(x^(2)).(1)/(((y-x))/(x))`
`= ((1+logy)^(2))/(logy)[:' logy = (y-x)/(x)logy = (y)/(x)-1rArr 1+ logy = (y)/(x)]`