Question

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

Answer 1

$\begin{array}{c}x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}}y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{\frac{\sqrt{cos2t}3{\cos }^{2}t\left(-\sin t\right)+{\cos }^{3}t\frac{2\sin 2t}{2\sqrt{\cos 2t}}}{\cos 2t}}{\frac{\sqrt{\cos 2t}3{\sin }^{2}t\cos t+{\sin }^{3}t\frac{2\sin t}{2\sqrt{\cos 2t}}}{\cos 2t}}\\ =\frac{-3{\cos }^{2}t\sin t\left(\cos 2t\right)+\sin 2t{\cos }^{3}t}{3{\sin }^{2}t\left(3\cos t\cos 2t+2{\sin }^{2}t\cos t\right)}\\ =\frac{{\cos }^{2}t\left(-3\sin t\cos 2t+2\sin t{\cos }^{2}t\right)}{{\sin }^{2}t\left(3\cos t\cos 2t+2{\sin }^{2}t\right)\cos t}\\ =\frac{{\cos }^{2}t\sin t\left(-3\cos 2t+2{\cos }^{2}t\right)}{{\sin }^{2}t\cos t\left(3\cos 2t+2{\sin }^{2}t\right)}\\ =\frac{\cos t\left(-6{\cos }^{2}t+3+2{\cos }^{2}t\right)}{\sin t\left(3-6{\sin }^{2}t+2{\sin }^{2}t\right)}\\ =\frac{3\cos t-4{\cos }^{3}t}{3\sin t-4{\sin }^{3}t}\\ =-\frac{\cos 3t}{\sin 3t}\\ =-\cot 3t.\end{array}$