I'm self teaching and stuck on the last question of the exercises on implicit differentiation. It says given that $\sin y=2\sin x$, show $(\frac{dy}{dx})^2=1 + 3\sec^2y$
My workings follow. I differentiate both sides w.r.t $x$, square and rearrange:
$$\cos y \frac{dy}{dx} = 2\cos x \Rightarrow \cos^2y(\frac{dy}{dx})^2 = 4\cos^2x$$
$$\Rightarrow (\frac{dy}{dx})^2 = \frac{4\cos^2x}{\cos^2y}$$
I'm now trying to rearrange the RHS to look like $1 + 3\sec^2y$ but failing. By employing the identity $\cos^2x + sin^2x = 1$and looking at the original equation, I can get to $\cos^2x = 1 – (\dfrac{\sin y}{2})^2$ and end up with
$$(\frac{dy}{dx})^2 = \frac{4(1 – \frac{1}{4}\sin^2y)}{\cos^2y} = \frac{4 – \sin^2y}{\cos^2y}$$
Can someone please put me on the right path? I wonder if I should differentiate both sides of $\cos y \frac{dy}{dx} = 2\cos x$ as that also leads to an equation involving $(\frac{dy}{dx})^2$.
Best Answer
Hint: use $4-\sin^2 y = 3+\cos^2 y$.