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

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

& \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{aligned}$$ Therefore, the differentiation of $y=\log {{x}^{x}}$ is $$\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.