Question

If $ y = {x^{1/x}}, $ the value of $ \dfrac{{dy}}{{dx}} $ at $ x = e $ is
A. 1
B. 0
C. -1
D. None of these

Answer 1

Hint: This problem deals with differentiation and logarithms. Given a complex function, where the exponent is a variable, so is the base. Hence this problem should be solved in a different way, rather than just differentiating the function directly.
Here basic differentiation formulas and chain rule in differentiation are used such as:
 $ \Rightarrow \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x} $
 $ \Rightarrow \dfrac{d}{{dx}}\left( {{f_1}(x).{f_2}(x)} \right) = {f_1}(x).\dfrac{d}{{dx}}\left( {{f_2}(x)} \right) + {f_2}(x).\dfrac{d}{{dx}}\left( {{f_1}(x)} \right) $

Complete step-by-step answer:
Given an equation which is a function of $ x $ , which is given by:
 $ \Rightarrow y = {x^{1/x}} $
We have to differentiate the above equation and after finding the derivative of the equation, we have to substitute $ e $ in the value of $ x $ , as given below:
We cannot differentiate the above equation directly, but we can proceed with differentiation after applying logarithms on both sides of the equation.
Consider the equation $ y = {x^{1/x}} $ , and apply logarithms on both sides, as given below:
 $ \Rightarrow y = {x^{1/x}} $
 $ \Rightarrow {\log _e}y = {\log _e}{x^{1/x}} $
 \[ \Rightarrow {\log _e}y = \dfrac{1}{x}{\log _e}x\]
Now differentiating the above equation with respect to $ x $ on both sides, as given below:
 $ \Rightarrow \dfrac{d}{{dx}}\left( {{{\log }_e}y} \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{x}{{\log }_e}x} \right) $
Applying the chain rule of differentiation to the R.H.S of the above equation:
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\dfrac{d}{{dx}}\left( {{{\log }_e}x} \right) + {\log _e}x\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) $
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\left( {\dfrac{1}{x}} \right) + {\log _e}x\left( {\dfrac{{ - 1}}{{{x^2}}}} \right) $
We know that derivative of $ {\log _e}x = \dfrac{1}{x} $ and the derivative of $ \dfrac{1}{x} $ is $ - \dfrac{1}{{{x^2}}} $ .
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^2}}}{\log _e}x $
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^2}}}\left( {1 - {{\log }_e}x} \right) $
Obtaining the expression of $ \dfrac{{dy}}{{dx}} $ , as given below:
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\left( {1 - {{\log }_e}x} \right) $
Now substitute $ x = e $ in the above equation, as given below:
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{{e^2}}}\left( {1 - {{\log }_e}e} \right) $
We know that the value of logarithm of $ e $ to the base $ e $ is equal to 1, which is expressed below:
 $ \Rightarrow {\log _e}e = 1 $
Substituting this value in the obtained equation, gives:
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{{e^2}}}\left( {1 - 1} \right) $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{{e^2}}}\left( 0 \right) $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = 0 $

Final Answer: The value of $ \dfrac{{dy}}{{dx}} $ at $ x = e $ is 0.

Note:
Please note that while solving the problem, when given the function of $ y = {x^{1/x}} $ , and we are asked to find $ \dfrac{{dy}}{{dx}} $ , then we cannot just simply differentiate the given function, because the function is complex. Whenever such kind of complex functions are given, apply logarithms to the given function and then find the differentiation of the applied logarithmic function, which would be easy to solve. We know that $ {\log _a}a = 1 $ , which is a basic logarithmic formula, is applied here.