Question

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

Answer 1

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.