CUET Preparation Today
| Given $x=\sin t$ and $y=\cos 2t$. What is $\frac{dy}{dx}$? |
$x$ $4x$ $-4x$ $x^2$ |
| $-4x$ |
| We have $\frac{dx}{dt}=\cos t$ and $\frac{dy}{dt}=-2\sin 2t$. Hence $\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=-2\sin 2t/\cos t=-4\sin t\cos t/\cos t=-4\sin t=-4x$ |
