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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.

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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 $$\begin{align} \frac{d^2y}{dx^2}&=\frac{-\sin(y)\cos(y)+\tan(y)(3+\cos^2(y))}{\cos^2(y)}\\ &=\frac{-\sin(y)\cos(y)+3\tan(y)+\sin(x)\cos(y)}{\cos^2(y)}\\ &=3\sec^2(y)\tan(y) \end{align}$$

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  • $\begingroup$ Ahh okay! This makes a lot of sense, i was wondering that when you manipulated the identity how you knew which root to take? Like why did you do the positive root instead of the negative? $\endgroup$ – user639649 Jan 31 '19 at 20:05
  • $\begingroup$ I am not sure if I totally understand your question. It doesn't really matter whether I take positive or negative root. When I square both sides, they will become positive. $\endgroup$ – Larry Jan 31 '19 at 20:25

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