Question-142045

Question Number 142045 by Engr_Jidda last updated on 25/May/21

Answered by Dwaipayan Shikari last updated on 25/May/21

y=((1/x))^x log(y)=−xlog(x)⇒((y′)/y)=−1−log(x)⇒y=−x^(−x) (1+log(x))

$${y}=\left(\frac{\mathrm{1}}{{x}}\right)^{{x}} \\ $$$${log}\left({y}\right)=−{xlog}\left({x}\right)\Rightarrow\frac{{y}'}{{y}}=−\mathrm{1}−{log}\left({x}\right)\Rightarrow{y}=−{x}^{−{x}} \left(\mathrm{1}+{log}\left({x}\right)\right) \\ $$

Commented by Engr_Jidda last updated on 25/May/21

$${thanks} \\ $$

Answered by EDWIN88 last updated on 26/May/21

(1) y=log (cos x) ⇒cos x = e^y ⇒ −sin x = e^y . y′ ⇒y′=−sin x.e^(−y) = −sin x.e^(−log (cos x)) ⇒y′=−sin x.sec x

$$\left(\mathrm{1}\right)\:\mathrm{y}=\mathrm{log}\:\left(\mathrm{cos}\:\mathrm{x}\right)\:\Rightarrow\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{e}^{\mathrm{y}} \\ $$$$\Rightarrow\:−\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{e}^{\mathrm{y}} .\:\mathrm{y}' \\ $$$$\Rightarrow\mathrm{y}'=−\mathrm{sin}\:\mathrm{x}.\mathrm{e}^{−\mathrm{y}} \:=\:−\mathrm{sin}\:\mathrm{x}.\mathrm{e}^{−\mathrm{log}\:\left(\mathrm{cos}\:\mathrm{x}\right)} \\ $$$$\Rightarrow\mathrm{y}'=−\mathrm{sin}\:\mathrm{x}.\mathrm{sec}\:\mathrm{x}\: \\ $$

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Question-142045

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