Differentiation Questions

Question Number 102342 by bemath last updated on 08/Jul/20

$${If}\:\begin{cases}{{x}=\mathrm{2}{t}+\mathrm{sin}\:\mathrm{2}{t}}\\{{y}={e}^{\mathrm{sin}\:\mathrm{2}{t}} }\end{cases} \\$$$${prove}\:{that}\:\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dx}}\:=\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−{t}\right) \\$$

Commented by Dwaipayan Shikari last updated on 08/Jul/20

$${There}\:{is}\:{some}\:{error}\:{in}\:{question}\:\:{if}\:\:\:{x}=\mathrm{2}{t}−{cos}\mathrm{2}{t}\:\:{then}\:{the}\:{prove}\:{is}\:{true} \\$$

Commented by bobhans last updated on 08/Jul/20

$${yes}\:{sir}.\:{i}\:{think}\:{something}\:{error}\:{in}\:{question} \\$$

Answered by bobhans last updated on 08/Jul/20

$$\Rightarrow\mathrm{sin}\:\mathrm{2}{t}\:=\:\mathrm{ln}\left(\mathrm{y}\right)\:;\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{t}\:=\:\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dt}} \\$$$$\frac{{dx}}{{dt}}\:=\:\mathrm{2}+\mathrm{2cos}\:\mathrm{2}{t}\:\Rightarrow\frac{{dx}}{{dt}}\:=\:\mathrm{2}+\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dt}}\:...\left(\mathrm{1}\right) \\$$$$\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{\mathrm{1}}{\frac{{dx}}{{dt}}}\: \\$$$$\frac{{dy}}{{dx}}=\:\mathrm{2}{y}\:\mathrm{cos}\:\mathrm{2}{t}\:×\:\frac{\mathrm{1}}{\mathrm{2}+\mathrm{2cos}\:\mathrm{2}{t}} \\$$$$\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2cos}\:\mathrm{2}{t}}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{t}\right)}\:=\:\frac{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {t}}{\mathrm{2cos}\:^{\mathrm{2}} {t}}\: \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} {t}−\mathrm{tan}\:^{\mathrm{2}} {t}\: \\$$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}\:^{\mathrm{2}} {t}−\mathrm{1}\right)−\mathrm{tan}\:^{\mathrm{2}} {t} \\$$$$\\$$

Answered by Dwaipayan Shikari last updated on 08/Jul/20

$${if}\:\:{x}=\mathrm{2}{t}−{cos}\mathrm{2}{t} \\$$$$\frac{{dx}}{{dt}}=\mathrm{2}+\mathrm{2}{sin}\mathrm{2}{t} \\$$$${logy}={sin}\mathrm{2}{t} \\$$$$\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dt}}=\mathrm{2}{cos}\mathrm{2}{t} \\$$$$\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dx}}=\frac{\mathrm{2}{cos}\mathrm{2}{t}}{\mathrm{2}+\mathrm{2}{sin}\mathrm{2}{t}}=\frac{\frac{\mathrm{1}−{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}{\mathrm{1}+\frac{\mathrm{2}{tant}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}=\frac{\mathrm{1}−{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{tant}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−{tant}}{\mathrm{1}+{tant}}={tan}\left(\frac{\pi}{\mathrm{4}}−{t}\right) \\$$