Question

Find $d=\frac{dy}{dx}$ if $x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}}$
$y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$

Answer 1

$x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}}$

$\Rightarrow \frac{dx}{dt}=\frac{\sqrt{\cos 2t}.\frac{d}{dt}{\sin }^{3}t-{\sin }^{3}t.\frac{d}{dt}\sqrt{\cos 2t}}{{\left(\sqrt{\cos 2t}\right)}^2}$

$\Rightarrow \frac{dx}{dt}=\frac{3{\sin }^{2}t.\cos t.\cos 2t+{\sin }^{3}t.\sin 2t}{{\left(\cos 2t\right)}^{\frac{3}{2}}}$

$and,y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$

$\Rightarrow \frac{dy}{dt}=\frac{\sqrt{\cos 2t}.\frac{d}{dt}{\cos }^{3}t-{\cos }^{3}t.\frac{d}{dt}\sqrt{\cos 2t}}{{\left(\sqrt{\cos 2t}\right)}^2}$

$\Rightarrow \frac{dy}{dt}=\frac{-3\sin t.{\cos }^{2}t.\cos 2t+{\cos }^{3}t.\sin 2t}{{\left(\cos 2t\right)}^{\frac{3}{2}}}$

$\because \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\Rightarrow \frac{dy}{dx}=\frac{{\cos }^{3}t.\sin 2t-3\sin t.{\cos }^{2}t.\cos 2t}{3{\sin }^{2}t.\cos t.\cos 2t+{\sin }^{3}t.\sin 2t}$

$\Rightarrow \frac{dy}{dx}=\frac{2{\cos }^{3}t-3\cos t.\cos 2t}{3\sin t.\cos 2t+2{sin}^{3}t}(\because \sin 2\theta =2\sin \theta .\cos \theta )$