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Question
If xpyq = (x + y)p+q then Prove that `dy/dx = y/x`
Solution
xpyq = (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`
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