Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com