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.
Im not sure where I've gone wrong, please help.
Thank you.
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 $$\begin{align} \frac{d^2y}{dx^2}&=\frac{-\sin(y)\cos(y)+\tan(y)(3+\cos^2(y))}{\cos^2(y)}\\ &=\frac{-\sin(y)\cos(y)+3\tan(y)+\sin(x)\cos(y)}{\cos^2(y)}\\ &=3\sec^2(y)\tan(y) \end{align}$$