Question

Use logarithmic differentiation to find dy/dx. y = (1 + x)^1/x, x > 0

Answer 1

Use exp/log functions to rewrite y as

`y = (1+x)^(\frac1x) \implies y = \exp\left(\ln\left(1+x)^(\frac1x)\right)\right)`

One of the properties of logarithms lets us bring down the exponent, so

`y = \exp\left(\frac{\ln(1+x)}x\right)`

Now take the derivative of both sides with respect to x. By the chain and quotient rules,

`(dy)/(dx) = \exp\left(\frac{\ln(1+x)}x\right) * \frac{\frac x{1+x} - \ln(1+x)}{x^2}`

Simplify:

`\boxed{(dy)/(dx) = (x - (1+x)\ln(1+x))/(x^2(1+x)) (1+x)^(\frac1x)}`