Question

Differentiate $\dfrac{{\tan x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)$

Answer 1

Hint : We can find the differentiation of $\dfrac{{\tan x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)$ using the product rule. Product rule is a rule to differentiate problems where one function is multiplied by another function. We cannot directly differentiate a product of functions so we separate and differentiate the functions according to the product rule. First simplify $\dfrac{{\tan x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)$ into a simpler expression by using appropriate logarithmic rules and then apply the product rule.
Formula used:
According to the product rule, $\dfrac{d}{{dx}}\left( {uv} \right) = u\left( {\dfrac{{dv}}{{dx}}} \right) + v\left( {\dfrac{{du}}{{dx}}} \right)$

Complete step-by-step answer :
We are given find the derivative of $\dfrac{{\tan x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)$ with respect to x.
$\dfrac{d}{{dx}}\left[ {\dfrac{{\tan x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)} \right]$
We have, $\log \dfrac{{{e^x}}}{{{x^x}}}$ which can also be written as $\log {\left( {\dfrac{e}{x}} \right)^x}$ which is equal to $x\log \left( {\dfrac{e}{x}} \right)$ as $\log {a^m} = m\log a$ .
This gives $x\log \left( {\dfrac{e}{x}} \right) = x\left( {\log e - \log x} \right) = x\left( {1 - \log x} \right)$ as $\log \dfrac{a}{b} = \log a - \log b$ and the value of $\log e$ is 1.
So $\dfrac{{\tan x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)$ becomes $\dfrac{{\tan x}}{x} \times x\left( {1 - \log x} \right) = \tan x\left( {1 - \log x} \right)$
So now we have to find the derivative of $\dfrac{d}{{dx}}\left[ {\tan x\left( {1 - \log x} \right)} \right]$
Considering $\tan x$ as u and $\left( {1 - \log x} \right)$ as v, $\dfrac{d}{{dx}}\left[ {\tan x\left( {1 - \log x} \right)} \right]$ becomes $\dfrac{d}{{dx}}\left( {uv} \right)$ which is a product.
According to the product rule, $\dfrac{d}{{dx}}\left( {uv} \right) = u\left( {\dfrac{{dv}}{{dx}}} \right) + v\left( {\dfrac{{du}}{{dx}}} \right)$
Therefore, $\dfrac{d}{{dx}}\left[ {\tan x\left( {1 - \log x} \right)} \right]$ becomes $\tan x\dfrac{d}{{dx}}\left( {1 - \log x} \right) + \left( {1 - \log x} \right)\dfrac{d}{{dx}}\tan x$
Derivative of $\tan x$ is $se{c^2}x$ and the derivative of $\log x$ is $\dfrac{1}{x}$
Substituting these values in $\tan x\dfrac{d}{{dx}}\left( {1 - \log x} \right) + \left( {1 - \log x} \right)\dfrac{d}{{dx}}\tan x$ , we get $\tan x\left( {0 - \dfrac{1}{x}} \right) + \left( {1 - \log x} \right)\left( {se{c^2}x} \right) = se{c^2}x\left( {1 - \log x} \right) - \dfrac{{\tan x}}{x}$
Therefore, the differentiation of $\dfrac{{\tan x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)$ is $se{c^2}x\left( {1 - \log x} \right) - \dfrac{{\tan x}}{x}$
So, the correct answer is “$se{c^2}x\left( {1 - \log x} \right) - \dfrac{{\tan x}}{x}$ ”.

Note : Do not confuse between differentiation and integration. For example, differentiation of sine gives positive cosine whereas integration of sine gives negative cosine. As we can see, except for the sign, the whole result is the same. So be careful with the results of differentiations and integration and while applying the product rule as we may get confused u with v and v with u.