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

$y = \exp\left(\dfrac{\ln(1+x)}x\right)$

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

$\dfrac{dy}{dx} = \exp\left(\dfrac{\ln(1+x)}x\right) \times \dfrac{\frac x{1+x} - \ln(1+x)}{x^2}$

Simplify:

$\boxed{\dfrac{dy}{dx} = \frac{x - (1+x)\ln(1+x)}{x^2(1+x)} (1+x)^{\frac1x}}$