Question Number 64872 by Rio Michael last updated on 22/Jul/19
$${Given}\:{that}\: \\ $$$$\:{y}\:=\:\left({cosx}^{} \right)^{{sinx}} \:\:{find}\:\frac{{dy}}{{dx}} \\ $$$${and}\: \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{y} \\ $$
Commented by kaivan.ahmadi last updated on 22/Jul/19
$${lny}={sinxln}\left({cosx}\right)\Rightarrow\frac{{y}'}{{y}}={cosxln}\left({cosx}\right)−\frac{{sinx}}{{cosx}}{sinx}\Rightarrow \\ $$$$\frac{{dy}}{{dx}}=\left({cosx}\right)^{{sinx}} \left({cosxln}\left({cosx}\right)−\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}\right) \\ $$$$ \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} {cosx}^{{sinx}} =\mathrm{1}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 22/Jul/19
$${we}\:{have}\:{y}\:={e}^{{sinxln}\left({cosx}\right)} \:\Rightarrow\frac{{dy}}{{dx}}\:=\frac{{d}}{{dx}}\left(\:{sinx}\:{ln}\left({cosx}\right)\right){e}^{{sinxln}\left({cosx}\right)} \\ $$$$=\left\{{cosxln}\left({cosx}\right)\:+{sinx}\:\frac{−{sinx}}{{cosx}}\right\}{y}\left({x}\right)\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)=\frac{{cos}^{\mathrm{2}} {x}\:{ln}\left({cosx}\right)−{sin}^{\mathrm{2}} {x}}{{cosx}}×\:\left({cosx}\right)^{{sinx}} \\ $$
Commented by mathmax by abdo last updated on 22/Jul/19
$${we}\:{have}\:{y}\left({x}\right)\:={e}^{{sinxln}\left({cosx}\right)} \:\:\:\:{and}\:{lim}_{{x}\rightarrow\mathrm{0}} {sinxln}\left({cosx}\right)\:=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{y}\left({x}\right)\:=\mathrm{1} \\ $$
Commented by Masumsiddiqui399@gmail.com last updated on 23/Jul/19
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