Question

If $y=\log {{x}^{x}}$, then the value $\dfrac{dy}{dx}$ is
A. ${{x}^{x}}\left( 1+\log x \right)$
B. $\log \left( ex \right)$
C. $\log \left( \dfrac{e}{x} \right)$
D. $\log \left( \dfrac{x}{e} \right)$

Answer 1

Hint: We first define the multiplication rule and how the differentiation of function works. We take the addition of these two different differentiated values. We take the $\dfrac{dy}{dx}$ altogether. We keep one function and differentiate the other one and then do the same thing with the other function. Then we take the addition to complete the formula.

Complete answer:
We now discuss the multiplication process of two functions where \[f\left( x \right)=u\left( x \right)v\left( x \right)\]
Differentiating \[f\left( x \right)=uv\], we get \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\].
We simplify the logarithm by $y=\log {{x}^{x}}=x\log x$.
The above-mentioned rule is the multiplication rule. We apply that on $f\left( x \right)=x\log x$. We assume the functions where \[u\left( x \right)=x,v\left( x \right)=\log x\]
We know that differentiation of \[u\left( x \right)=x\] is ${{u}^{'}}\left( x \right)=1$ and differentiation of \[v\left( x \right)=\log x\] is \[{{v}^{'}}\left( x \right)=\dfrac{1}{x}\]. We now take differentiation on both parts of $f\left( x \right)=x\log x$ and get \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ x\log x \right]\].
We place the values of ${{u}^{'}}\left( x \right)=1$ and \[v\left( x \right)=\log x\] to get
\[\dfrac{d}{dx}\left[ x\log x \right]=x\dfrac{d}{dx}\left( \log x \right)+\left( \log x \right)\dfrac{d}{dx}\left( x \right)\].
We take all the $\dfrac{dy}{dx}$ forms altogether to get
\[\begin{align}
  & \dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ x\log x \right] \\
 & \Rightarrow {{f}^{'}}\left( x \right)=x\times \dfrac{1}{x}+\left( \log x \right)\left( 1 \right) \\
 & \Rightarrow {{f}^{'}}\left( x \right)=1+\log x=\log e+\log x \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\log \left( ex \right) \\
\end{align}\]
Therefore, the differentiation of $y=\log {{x}^{x}}$ is \[\log \left( ex \right)\].
And hence the correct answer is option B.

Note:
We need remember that in the chain rule \[\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}\], we aren’t cancelling out the part \[d\left[ h\left( x \right) \right]\]. Canceling the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate. The rule may be extended or generalized to many other situations, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts.