How do you differentiate ${x^{\dfrac{1}{x}}}$?

Hint:In this question we need to find the derivative of ${x^{\dfrac{1}{x}}}$. We can find the derivative of ${x^{\dfrac{1}{x}}}$ using chain rule, properties of logarithm and exponential function. Chain rule of differentiability is used for finding the derivative of a composite function such as $f(g(x)$ where both functions are differentiable.

Complete step by step solution:
 Let us try to find the derivative of${x^{\dfrac{1}{x}}}$.
Before finding the derivative of${x^{\dfrac{1}{x}}}$. We have to first know, what is the chain rule of differentiability?
In chain rule we have a real valued composite of two functions $f$ such that $f(x)\, = \,v(u(x))$ such that both functions are differentiable. Suppose $t = u(x)$
$\dfrac{{d(f(x)}}{{dx}}\, = \,\dfrac{{d(v(u(x))}}{{dx}} = \dfrac{{d(v(t)}}{{dx}} =
\,\dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$
Our function which we have to differentiate can be written as this by using the property of exponential and logarithm function.
As we know that${e^{\ln \,x}}\,\, = \,x$. So by using this, we get
${x^{\dfrac{1}{x}}}\, = \,{e^{\ln \,{x^{\dfrac{1}{x}}}}}$ And also as we know$\ln \,{a^b} = b \cdot
\ln \,a$.
 Using this rule we get
${x^{\dfrac{1}{x}}}\, = \,{e^{\dfrac{1}{x}\,\ln \,x}}$
Now let us find the derivative by applying chain rule.
\[\dfrac{{d({x^{\dfrac{1}{x}}})}}{{dx}}\, = \,\dfrac{{d({e^{\dfrac{1}{x}\,\ln x}})}}{{dx}} =
{e^{\dfrac{{\ln x}}{x}}} \cdot \dfrac{{d\left( {\dfrac{{\ln x}}{x}} \right)}}{{dx}}\] $eq(1)$
As we know that derivative of$\dfrac{{d({e^x})}}{{dx}} = {e^x}$.
Now we will use the division rule of derivatives. In division rule of derivative, we find derivative of two function such as
$\dfrac{{d(\dfrac{v}{u})}}{{dx}} =
\dfrac{{u\dfrac{{dv}}{{dx}} -
v\dfrac{{du}}{{dx}}}}{{{u^2}}}$.
So by using division rule we got derivative of
\[
\dfrac{{d\left( {\dfrac{{\ln x}}{x}} \right)}}{{dx}} = \dfrac{{x\dfrac{{d(\ln x)}}{{dx}} - \ln
x(\dfrac{{d(x)}}{{dx}})}}{{{x^2}}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\dfrac{{x \cdot \dfrac{1}{x} - \ln x}}{{{x^2}}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{1 - \ln x}}{{{x^2}}} \\
\]
Now putting the derivative of$\dfrac{{d\left( {\dfrac{{\ln x}}{x}} \right)}}{{dx}}$. We get the derivative of ${x^{\dfrac{1}{x}}}$
$

\dfrac{{d\left( {{x^{\dfrac{1}{x}}}} \right)}}{{dx}}\, = \,{e^{\dfrac{{\ln x}}{x}}}\left( {\dfrac{{1 - \ln
x}}{{{x^2}}}} \right) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^{\dfrac{1}{x}}}\left( {\dfrac{{1 - \ln x}}{{{x^2}}}} \right)
\\
$
Because we know ${x^{\dfrac{1}{x}}}\, = \,{e^{\dfrac{1}{x}\,\ln \,x}}$

Hence the derivative of ${x^{\dfrac{1}{x}}}$is${x^{\dfrac{1}{x}}}\left( {\dfrac{{1 - \ln x}}{{{x^2}}}} \right)$.

Note: In the question which asked to find the derivative of function you must need to know the rule of derivative such as Addition, subtraction, multiplication, division and chain rule without you not able to answer these questions. Also you have to know the definition of differentiability.