# How do you differentiate y=(1+x)^(1/x)?

Mar 31, 2017

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

#### Explanation:

We have:

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

First note that the function is not defined when $x = 0$. We can take (natural) logarithms of both sides to get:

$\ln y = \ln \left\{{\left(1 + x\right)}^{\frac{1}{x}}\right\}$
$\text{ } = \frac{1}{x} \setminus \ln \left(1 + x\right)$
$\text{ } = \ln \frac{1 + x}{x}$

Differentiating implicitly and applying the quotient rule and the chain rule gives:

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x\right) \left(\frac{1}{1 + x}\right) - \left(\ln \left(1 + x\right)\right) \left(1\right)}{x} ^ 2$
$\text{ } = \frac{\frac{x}{1 + x} - \ln \left(1 + x\right)}{x} ^ 2$
$\text{ } = \frac{1}{x \left(1 + x\right)} - \ln \frac{1 + x}{x} ^ 2$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left\{\frac{1}{x \left(1 + x\right)} - \ln \frac{1 + x}{x} ^ 2\right\}$
$\text{ } = {\left(1 + x\right)}^{\frac{1}{x}} \left\{\frac{1}{x \left(1 + x\right)} - \ln \frac{1 + x}{x} ^ 2\right\}$

Mar 31, 2017

The answer is $= {\left(1 + x\right)}^{\frac{1}{x}} \left(\frac{1}{x \left(1 + x\right)} - \ln \frac{1 + x}{x} ^ 2\right)$

#### Explanation:

We diiferentiate using logs.

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

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

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

Differentiating

$\left(\ln y\right) ' = \left(\frac{1}{x} \cdot \ln \left(1 + x\right)\right) '$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} \cdot \frac{1}{1 + x} - \frac{1}{x} ^ 2 \cdot \ln \left(1 + x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{1}{x \left(1 + x\right)} - \frac{1}{x} ^ 2 \ln \left(1 + x\right)\right)$

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