Question Number 114654 by mohammad17 last updated on 20/Sep/20 | ||
Commented by mohammad17 last updated on 20/Sep/20 | ||
$${help}\:{me}\:{sir} \\ $$ | ||
Commented by mohammad17 last updated on 20/Sep/20 | ||
$${thank}\:{hou}\:{sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{other}\:{question} \\ $$ | ||
Answered by bobhans last updated on 20/Sep/20 | ||
$$\left[\mathrm{2}\right]\:{x}={a}\mathrm{cos}^{\mathrm{3}} \:\theta\:\Rightarrow\frac{{dx}}{{d}\theta}\:=\:−\mathrm{3}{a}\mathrm{cos}\:^{\mathrm{2}} \theta\:\mathrm{sin}\:\theta\: \\ $$$$\:\:\:\:\:\:\:{y}={a}\mathrm{sin}\:^{\mathrm{3}} \theta\Rightarrow\frac{{dy}}{{d}\theta}\:=\:\mathrm{3}{a}\mathrm{sin}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{d}\theta}\:×\:\frac{{d}\theta}{{dx}}\:=\:\frac{\mathrm{3}{a}\mathrm{sin}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:\theta}{−\mathrm{3}{a}\mathrm{cos}\:^{\mathrm{2}} \theta\:\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\:\frac{{dy}}{{dx}}\:=\:−\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\:=\:−\mathrm{tan}\:\theta \\ $$$$\:\:\:\:\:\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}\:=\:\mathrm{sec}\:\theta \\ $$ | ||
Answered by Dwaipayan Shikari last updated on 20/Sep/20 | ||
$${x}^{{y}} ={y}^{{x}} \\ $$$${ylogx}={xlogy} \\ $$$$\frac{{y}}{{x}}+{logx}\frac{{dy}}{{dx}}={logy}+\frac{{x}}{{y}}.\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\left({logx}−\frac{{x}}{{y}}\right)={logy}−\frac{{y}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{logy}−\frac{{y}}{{x}}}{{logx}−\frac{{x}}{{y}}} \\ $$ | ||