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

Apr 1, 2018

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

#### Explanation:

When dealing with a function raised to the power of a function, logarithmic differentiation becomes necessary.

Let $y = {x}^{\frac{1}{x}}$

Then,

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

Recalling that $\ln \left({x}^{a}\right) = a \ln x :$

$\ln y = \frac{1}{x} \ln x$

$\ln y = \ln \frac{x}{x}$

Now, differentiate both sides with respect to $x ,$ meaning that the left side will be implicitly differentiated:

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \ln x}{x} ^ 2$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}} :$

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

Write everything in terms of $x :$

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

Apr 2, 2018

#### Explanation:

A bit of formula here:

$\frac{d}{\mathrm{dx}} \left[f {\left(x\right)}^{g \left(x\right)}\right] = f {\left(x\right)}^{g \left(x\right)} \cdot \frac{d}{\mathrm{dx}} \left[\ln \left(f \left(x\right)\right) \cdot g \left(x\right)\right]$

$\implies \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{x}}\right] = {x}^{\frac{1}{x}} \cdot \frac{d}{\mathrm{dx}} \left[\ln \left(x\right) \cdot \frac{1}{x}\right]$

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

$\implies \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{x}}\right] = {x}^{\frac{1}{x}} \cdot \left(\frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right] \cdot \frac{1}{x} + \ln \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left[\frac{1}{x}\right]\right)$

Some rules here:

$\frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right] = \frac{1}{x}$

$\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ if $n$ is a constant.

$\implies \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{x}}\right] = {x}^{\frac{1}{x}} \cdot \left(\frac{1}{x} \cdot \frac{1}{x} + \ln \left(x\right) \cdot - 1 \cdot {x}^{- 1 - 1}\right)$

$\implies \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{x}}\right] = {x}^{\frac{1}{x}} \left(\frac{1}{{x}^{2}} + \ln \left(x\right) \cdot - \frac{1}{{x}^{2}}\right)$

$\implies \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{x}}\right] = {x}^{\frac{1}{x}} \left[\frac{1}{{x}^{2}} \left(1 - \ln \left(x\right)\right)\right]$

$\implies \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{x}}\right] = {x}^{\frac{1}{x}} \left[\frac{1 - \ln \left(x\right)}{{x}^{2}}\right]$