Question-134676

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

Answered by benjo_mathlover last updated on 06/Mar/21

(2)(d/dx) [ e^(xy) +3ln (xy)+cos (xy) ]= 1 ⇒(y+xy′)e^(xy) +3(((y+xy′)/(xy)))−(y+xy′)sin (xy)=1

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

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

3) e^(6y) =5+sinx ⇒6y=ln(5+sinx) ⇒y=(1/6)ln(5+sinx) ⇒ (dy/dx)=(1/6)((cosx)/(5+sinx)) =((cosx)/(30+6sinx))

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

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

9) f(x)=((cosx−(1/2))/(x−(π/3))) we do the changement x−(π/3)=t ⇒ f(x)=f((π/3)+t)=g(t) (x→(π/3)⇔t→0) g(t)=((cos(t+(π/3))−(1/2))/t)=(((1/2)cost−((√3)/2)sint−(1/2))/t) =−(1/2)((1−cost)/t)−((√3)/2)((sint)/t) ⇒g(t)∼−(1/2)((t^2 /2)/t)−((√3)/2) ⇒g(t)∼−(t/4)−((√3)/2) ⇒lim_(t→0) g(t)=−((√3)/2)=lim_(x→(π/3)) f(x)

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

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

f(θ)=((sinθ)/(π−θ)) we do the changement π−θ=x ⇒ f(θ)=((sin(π−x))/x) =((sinx)/x) =g(x) we have lim_(x→0) g(x)=1 ⇒ lim_(θ→π) g(θ)=1

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

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