Question

$x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}},y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$ then find $\frac{dy}{dx}$

Answer 1

$x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}},y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$
$\frac{dx}{dt}=\frac{d}{dx}\left[\frac{{\sin }^{3}t}{\sqrt{\cos 2t}}\right]$
$=\frac{d}{dx}\left({\sin }^{3}t\right)\frac{1}{\sqrt{\cos 2t}}+\frac{d}{dx}\left(\frac{1}{\sqrt{\cos 2t}}\right){\sin }^{3}t$ (by chain rule)
$=\frac{3{\sin }^{2}t\cos t}{\sqrt{\cos 2t}}+{\sin }^{3}t\left(\frac{-1}{2}\right){\left(\cos 2t\right)}^{-3/2}\times (-2\sin 2t)$
$\frac{dx}{dt}=\frac{3{\sin }^{2}t\cos t}{\sqrt{\cos 2t}}+\frac{{\sin }^{2}t\sin 2t}{{\left(\cos 2t\right)}^{3/2}}$
$\frac{dy}{dt}=\frac{d}{dt}\left(\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}\right)$
$=\frac{d}{dx}\left({\cos }^{3}t\right)\frac{1}{\sqrt{\cos 2t}}+\frac{d}{dx}\left(\frac{1}{\sqrt{\cos 2t}}\right){\cos }^{3}t$
$=\frac{3{\cos }^{2}t\left(\sin t\right)}{\sqrt{\cos 2t}}+(-\frac{1}{2}){\left(\cos 2t\right)}^{-3/2}\times (-2\sin 2t){\cos }^{3}t$
$=\frac{-3{\cos }^{2}t\sin t}{\sqrt{\cos 2t}}+\frac{{\cos }^{3}t\sin 2t}{{\left(\cos 2t\right)}^{3/2}}$
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$
$=\frac{\frac{-3{\cos }^{2}t\sin t}{\sqrt{\cos 2t}}+\frac{{\cos }^{3}t\sin 2t}{{\left(\cos 2t\right)}^{3/2}}}{\frac{3{\sin }^{2}t\cos t}{\sqrt{\cos 2t}}+\frac{{\sin }^{3}t\sin 2t}{{\left(\cos 2t\right)}^{3/2}}}$
$=\frac{-3{\cos }^{2}t\sin t\cos 2t+{\cos }^{3}t\sin 2t}{3{\sin }^{2}t\cos t\cos 2t+{\sin }^{3}t\sin 2t}$
$=\frac{-3{\cos }^{2}t\sin t\left(\cos 2t\right)+2{\cos }^{4}t\sin t}{3{\sin }^{2}t\cos t\cos 2t+2{\sin }^{4}t\cos t}$
$=\frac{2{\cos }^{3}t-3\cos t\cos 2t}{2{\sin }^{3}t+3\sin t\cos 2t}$
$=\frac{2{\cos }^{3}t-3\left(\cos t\right)(2{\cos }^{2}t-1)}{2{\sin }^{3}t+3\sin t(1-2{\sin }^{2}t)}=\cos \left(t\right)\left[\frac{3-4{\cos }^{2}t}{3-4{\sin }^{2}t}\right]$