Question

If x^py^q = (x + y)^p+q then Prove that `dy/dx = y/x`

Answer 1

x^py^q = (x + y)^p+q

Taking log both side

p log x + q log y = (p + q) log (x + y)

Differentiating w.r.t. x

`p/x + q/y dy/dx = (p + q)/(x + y) + ((p + q)/(x + y))dy/dx`

`q/ydy/dx - ((p + q)/(x + y)) dy/dx = (p + q)/(x + y) - p/x`

`(q/y - (p + q)/(x + y)) dy/dx = ((p + q)/(x + y) - p/x)`

`((qx - py)/y)dy/dx = ((qx - py)/x)`

`1/y dy/dx = 1/x`

`dy/dx = y/x`