# How do you find the derivative of (1+x)^(1/x)?

Feb 5, 2017

$f ' \left(x\right) = {\left(1 + x\right)}^{\frac{1}{x}} \left(\frac{1}{x \left(x + 1\right)} - \ln \frac{1 + x}{x} ^ 2\right)$

#### Explanation:

Consider:

$f \left(x\right) = {\left(1 + x\right)}^{\frac{1}{x}}$

The function is defined only for $\left(1 + x\right) > 0$ so we can take its logarithm:

$\ln \left(f \left(x\right)\right) = \ln \left({\left(1 + x\right)}^{\frac{1}{x}}\right)$

and using the properties of logarithms:

$\ln \left(f \left(x\right)\right) = \frac{1}{x} \ln \left(1 + x\right)$

Now differentiate both sides, using the chain rule at first member and the product rule at the second member:

$\frac{f ' \left(x\right)}{f} \left(x\right) = \frac{1}{x \left(x + 1\right)} - \ln \frac{1 + x}{x} ^ 2$

and since we know $f \left(x\right)$:

$f ' \left(x\right) = {\left(1 + x\right)}^{\frac{1}{x}} \left(\frac{1}{x \left(x + 1\right)} - \ln \frac{1 + x}{x} ^ 2\right)$

Alternatively we can write the function as:

${\left(1 + x\right)}^{\frac{1}{x}} = {\left({e}^{\ln \left(1 + x\right)}\right)}^{\frac{1}{x}} = {e}^{\ln \frac{1 + x}{x}}$

using the chain rule:

$\frac{d}{\mathrm{dx}} {e}^{\ln \frac{1 + x}{x}} = {e}^{\ln \frac{1 + x}{x}} \frac{d}{\mathrm{dx}} \ln \frac{1 + x}{x} = {\left(1 + x\right)}^{\frac{1}{x}} \frac{d}{\mathrm{dx}} \ln \frac{1 + x}{x}$

which is the same expression we had in the other way.