# Deriving $dy/dx = 2\cos x/\cos y$ given $\sin y=2\sin x$

My original question is to find the second derivative of $$\sin y=2\sin x$$

I derived it once got $$2\cos x/\cos y$$ which was correct but the second time did not get $$3\sec^2y\tan y$$ which is the answer.

You got $$\frac{dy}{dx}=2\frac{\cos(x)}{\cos(y)}$$ Recall the quotient rule, then the second derivative will be \begin{align} \frac{d^2y}{dx^2}&=2\frac{d}{dx}\frac{\cos(x)}{\cos(y)}\\ &=2\frac{-\sin(x)\cos(y)+\sin(y)\frac{dy}{dx}\cos(x)}{\cos^2(y)}\\ &=\frac{-\sin(y)\cos(y)+4\tan(y)\cos^2(x)}{\cos^2(y)} \end{align} We know that $$\sin(y)=2\sin(x)$$ Then $$\sqrt{1-\cos^2(y)}=2\sqrt{1-\cos^2(x)}$$ $$1-\cos^2(y)=4(1-\cos^2(x))$$ $$\cos^2(x)=\frac{3+\cos^2(y)}{4}$$ Then \begin{align} \frac{d^2y}{dx^2}&=\frac{-\sin(y)\cos(y)+\tan(y)(3+\cos^2(y))}{\cos^2(y)}\\ &=\frac{-\sin(y)\cos(y)+3\tan(y)+\sin(x)\cos(y)}{\cos^2(y)}\\ &=3\sec^2(y)\tan(y) \end{align}