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

Hint: In differentiation, when dealing with a function raised to the power of function, logarithmic differentiation becomes necessary. Therefore, always remember this tip in such types of questions on the topic. The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to change in its argument.

Complete step-by-step answer:
Follow the steps to solve the given question
Step 1) Let $ y = {x^{\dfrac{1}{x}}} $
Then, we see
 $ \log y = \log ({x^{\dfrac{1}{x}}}) $
Step 2) Recalling that $ \log ({x^a}) = a\log x $ (formula)
 $ \log y = \dfrac{1}{x}\log x $
 $ \log y = \dfrac{{\log x}}{x} $
Step 3) Now, differentiate both sides with respect to $ x $ , meaning that the left side will be implicitly differentiated as shown below:
Step 4) Solve for $ \dfrac{{dy}}{{dx}} $ :
 $ \dfrac{{dy}}{{dx}} = y\left( {\dfrac{{1 - \log x}}{{{x^2}}}} \right) $
Step 5) Write everything in terms of $ x $ as shown below:
 $ \dfrac{{dy}}{{dx}} = {x^{\dfrac{1}{x}}}\left( {\dfrac{{1 - \log x}}{{{x^2}}}} \right) $
So, the correct answer is “ $ \dfrac{{dy}}{{dx}} = {x^{\dfrac{1}{x}}}\left( {\dfrac{{1 - \log x}}{{{x^2}}}} \right) $ ”.

Note: In Calculus, differentiation (it is a method in which we find the instantaneous rate of change in a function based on one of its variables) is a method of finding the derivative of a function, like we did in the given question.
If $ x $ is a variable and $ y $ is another variable, then the rate of change of $ x $ with respect to $ y $ is given by $ \dfrac{{dy}}{{dx}} $ . This is the general expression of a function and is represented as $ f'(x) = \dfrac{{dy}}{{dx}} $ where $ y = f(x) $ is any function.