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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 $(\frac{dy}{dx})^2$.

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  • $\begingroup$ You need to put your final denominator in terms of $\cos y$ - then see how it looks. $\endgroup$ Apr 30, 2013 at 11:03

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Hint: use $4-\sin^2 y = 3+\cos^2 y$.

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  • $\begingroup$ Thanks to your hint, I get the right answer now. $\dfrac{3 + \cos^2y}{\cos^2y} = 1 + 3\sec^2y$ $\endgroup$
    – PeteUK
    Apr 30, 2013 at 11:09

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