Question

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

Answer 1

:A derivative is the rate at which output changes with respect to an input. We know that $\log \,{x^x} = x\,\,\log \,\,x$.

Complete step by step solution:
Let $y = \dfrac{{\tan x}}{x}\left( {\log \,\,\dfrac{{{e^x}}}{{{x^x}}}} \right)$
$y = \dfrac{{\tan x}}{x}\left( {\log \,\,\dfrac{{{e^x}}}{{{x^x}}}} \right)$
$y = \dfrac{{\tan x}}{x}\left( {\log \,\,{{\left( {\dfrac{e}{x}} \right)}^x}} \right)$
$y = \dfrac{{\tan x}}{x}\,x\, \times \,\log \,\,\left( {\dfrac{e}{x}} \right)$........................$\left( {\because \log \,{x^x} = x\,\,\log \,\,x} \right)$
$y = \tan x\,\,\left( {\log \,\,e\,\, - \,\,\log \,x} \right)$……………………..$\left[ {\because \log \left( {\dfrac{a}{b}} \right) = \log a - \log b} \right]$
$y = \tan x\,\,\left( {1\,\, - \,\,\log \,x} \right)$…...........................$\left( {\because \log e = 1} \right)$
$y = \tan x\,\, - \tan \,x\,\,\log \,x$
We will differentiate y with respect to x.
$\dfrac{{dy}}{{dx}} = {\sec ^2}x - \left[ {\tan x\dfrac{d}{{dx}}\log \,x\,\, + \,\,\log \,x\,\,\dfrac{d}{{dx}}\tan \,x} \right]$
When we differentiate$\tan \,x\,\,\log \,x$ then we 1st take${\text{tan}}\,{\text{x}}$as a constant term and differentiate${\text{log}}\,{\text{x}}$, further${\text{log}}\,{\text{x}}$ is a constant term and differentiate the value${\text{tan}}\,{\text{x}}$.

$$\dfrac{{dy}}{{dx}} = {\sec ^2}x - \left[ {\tan x \times \dfrac{1}{x} \times \dfrac{d}{{dx}}\,x\,\, + \,\,\log \,x\,\,\dfrac{d}{{dx}}\tan \,x} \right] \\\ \dfrac{{dy}}{{dx}} = {\sec ^2}x - \left[ {\tan x\dfrac{1}{x}x \times 1 + \log x \times {{\sec }^2}x} \right] \\\$$

$\dfrac{{dy}}{{dx}} = {\sec ^2}x - \dfrac{{\tan x}}{x}\,\, - \,\,\log \,\,x\,\,{\sec ^2}x$
We will take common${\sec ^2}x$, we will get
$\dfrac{{dy}}{{dx}} = {\sec ^2}x\left( {1 - \log x} \right) - \dfrac{{\tan x}}{x}\,\,$

Note: The properties of logarithm are:
(i) $\log {(a)^m} = m\log a$
(ii)$\log a.\log b = \log \left( {a + 3} \right)$
(iii)$\log \left( {\dfrac{a}{b}} \right) = \log a - \log b$
(iv)$\log 1 = 0$
(v)$\log e = 1$