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Question Number 134676 by mohammad17 last updated on 06/Mar/21

Answered by benjo_mathlover last updated on 06/Mar/21

$$\left(2\right)\frac{\mathrm{d}}{\mathrm{dx}}[{\mathrm{e}}^{\mathrm{xy}}+3\ln \left(\mathrm{xy}\right)+\cos \left(\mathrm{xy}\right)]=1$$$$\Rightarrow (\mathrm{y}+{\mathrm{xy}}^{\prime }){\mathrm{e}}^{\mathrm{xy}}+3\left(\frac{\mathrm{y}+{\mathrm{xy}}^{\prime }}{\mathrm{xy}}\right)-(\mathrm{y}+{\mathrm{xy}}^{\prime })\sin \left(\mathrm{xy}\right)=1$$$$$$

Answered by mathmax by abdo last updated on 06/Mar/21

$$3){\mathrm{e}}^{6\mathrm{y}}=5+\mathrm{sinx}\Rightarrow 6\mathrm{y}=\ln (5+\mathrm{sinx})\Rightarrow \mathrm{y}=\frac{1}{6}\ln (5+\mathrm{sinx})\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{6}\frac{\mathrm{cosx}}{5+\mathrm{sinx}}=\frac{\mathrm{cosx}}{30+6\mathrm{sinx}}$$

Answered by mathmax by abdo last updated on 06/Mar/21

$$9)\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{cosx}-\frac{1}{2}}{\mathrm{x}-\frac{\pi }{3}}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}-\frac{\pi }{3}=\mathrm{t}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\frac{\pi }{3}+\mathrm{t})=\mathrm{g}\left(\mathrm{t}\right)(\mathrm{x}\to \frac{\pi }{3}\iff \mathrm{t}\to 0)$$$$\mathrm{g}\left(\mathrm{t}\right)=\frac{\cos (\mathrm{t}+\frac{\pi }{3})-\frac{1}{2}}{\mathrm{t}}=\frac{\frac{1}{2}\mathrm{cost}-\frac{\sqrt{3}}{2}\mathrm{sint}-\frac{1}{2}}{\mathrm{t}}$$$$=-\frac{1}{2}\frac{1-\mathrm{cost}}{\mathrm{t}}-\frac{\sqrt{3}}{2}\frac{\mathrm{sint}}{\mathrm{t}}\Rightarrow \mathrm{g}\left(\mathrm{t}\right)\sim -\frac{1}{2}\frac{\frac{{\mathrm{t}}^2}{2}}{\mathrm{t}}-\frac{\sqrt{3}}{2}\Rightarrow \mathrm{g}\left(\mathrm{t}\right)\sim -\frac{\mathrm{t}}{4}-\frac{\sqrt{3}}{2}$$$$\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=-\frac{\sqrt{3}}{2}={\lim }_{\mathrm{x}\to \frac{\pi }{3}}\mathrm{f}\left(\mathrm{x}\right)$$

Answered by mathmax by abdo last updated on 06/Mar/21

$$\mathrm{f}\left(\theta \right)=\frac{\sin \theta }{\pi -\theta }\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\pi -\theta =\mathrm{x}\Rightarrow$$$$\mathrm{f}\left(\theta \right)=\frac{\sin (\pi -\mathrm{x})}{\mathrm{x}}=\frac{\mathrm{sinx}}{\mathrm{x}}=\mathrm{g}\left(\mathrm{x}\right)\mathrm{we}\mathrm{have}{\lim }_{\mathrm{x}\to 0}\mathrm{g}\left(\mathrm{x}\right)=1\Rightarrow$$$${\lim }_{\theta \to \pi }\mathrm{g}\left(\theta \right)=1$$$$$$

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