Question

If `x^y = e^(x-y), "prove that" dy/dx = logx/(1 + log x)^2`.

Answer 1

`x^y = e^(x-y)`

taking log on both sides

y log x = x − y log e   ...[∴ log e = 1]

y log x + y = x

`y = x/(log x + 1)`

Now we apply quotient rule,

`dy/dx = ((log x +1) d/dx(x) - x d/dx (log x + 1))/(1+log x)^2`

`= ((log x + 1)- x xx1/x)/(1+log x)^2`

`= (log x + 1-1)/(1+log x)^2`

`=logx/(1+logx)^2`

Hence proved

`dy/dx = logx/(1 + log x)^2`