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Question Number 69764 by Rio Michael last updated on 27/Sep/19
find   (dy/dx)  if  x = sin^2 t  and  y= tan t at  t = (π/4)
$${find}\:\:\:\frac{{dy}}{{dx}}\:\:{if}\:\:{x}\:=\:{sin}^{\mathrm{2}} {t}\:\:{and}\:\:{y}=\:{tan}\:{t}\:{at}\:\:{t}\:=\:\frac{\pi}{\mathrm{4}} \\ $$
Commented by kaivan.ahmadi last updated on 27/Sep/19
(dy/dx)=((dy/dt)/(dx/dt))=((1+tan^2 t)/(sin2t))
$$\frac{{dy}}{{dx}}=\frac{\frac{{dy}}{{dt}}}{\frac{{dx}}{{dt}}}=\frac{\mathrm{1}+{tan}^{\mathrm{2}} {t}}{{sin}\mathrm{2}{t}} \\ $$
Answered by mind is power last updated on 27/Sep/19
(dy/dx)=(dy/dt).(dt/dx)=(dy/dt).((dx/dt))^(−1) =(1+tg^2 (t)).(sin(2t))^(−1) ∣_(t=(π/4)) =2.1^(−1) =2
$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}.\frac{{dt}}{{dx}}=\frac{{dy}}{{dt}}.\left(\frac{{dx}}{{dt}}\right)^{−\mathrm{1}} =\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right).\left({sin}\left(\mathrm{2}{t}\right)\right)^{−\mathrm{1}} \mid_{{t}=\frac{\pi}{\mathrm{4}}} =\mathrm{2}.\mathrm{1}^{−\mathrm{1}} =\mathrm{2} \\ $$$$ \\ $$

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