Question

Find `dy/dx , if y^x = e^(x+y)`

Answer 1

`y^x = e^(x+y)` ...(Given)

Taking log on both sides, we get

`therefore log(y)^x = log(e)^(x+y)`

`therefore x.logy = x+ y .loge`

`therefore x log y = (x+y).1`

`therefore x. log y = x+y`

Differentiating w.r.t x, we get

`therefore x1/y dy/dx + log y . 1`

`=1dy/dx`

`therefore x1/y dy/dx + log y`

`=1+dy/dx`

`therefore dy/dx (x/y - 1) = 1- log y`

`therefore dy/dx = ((1-log y)(y))/(x-y)`