
Differentiate\[\dfrac{{\tan \,\,x}}{x}\left( {\log \dfrac{{{e^x}}}{{{x^x}}}} \right)\].
Answer
484.2k+ views
Hint::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)\[\log e = 1\]
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)\[\log e = 1\]
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