Question

The value of \[\dfrac{d}{{dx}}\left( {{x^x}} \right)\] is equal to:
A. \[{x^x}\log \left( {\dfrac{e}{x}} \right)\]
B. \[{x^x}\log ex\]
C. \[{x^x}\left( {1 + \log x} \right)\]
D. \[{x^x}\log x\]

Answer 1

Hint: First of all, apply logarithm to the function to obtain a simple equation. Then use the product rule of derivatives to find the derivative of the given function. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Let \[y = {x^x}\]
Applying logarithms on both sides, we get
\[ \Rightarrow \log y = \log {x^x}\]
We know that \[\log {a^b} = b\log a\]
\[ \Rightarrow \log y = x\log x\]
Differentiating on both sides w.r.t \[x\], we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {x\log x} \right)\]
By product rule of derivatives, we have
\[
   \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\left( {\log x} \right) + \log x\dfrac{d}{{dx}}\left( x \right) \\
   \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x \times 1 \\
   \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 1 + \log x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = y\left( {1 + \log x} \right) \\
  \therefore \dfrac{{dy}}{{dx}} = {x^x}\left( {1 + \log x} \right){\text{ }}\left[ {\because y = {x^x}} \right] \\
\]
Therefore, the derivative of \[{x^x}\] is \[{x^x}\left( {1 + \log x} \right)\].
Thus, the correct option is C. \[{x^x}\left( {1 + \log x} \right)\].

Note: The product rule states that if \[f\left( x \right)\] and \[g\left( x \right)\] are both differentiable, then \[\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]\]. Remember the derivative of \[{x^x}\] as a formula which will be useful to solve higher derivative problems.