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I had the question $9x^2 - y^2 = 1$ and my answer was $$\frac{\mathrm{d}x}{\mathrm{d}y} = -18x-2y .$$

I was wondering if I tackled this correctly? I am new to this.

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Wait wait its dy/dx = -18x / -2y right? – soniccool Oct 24 '11 at 23:45
I edited the title. I am not sure if you want to compute $\frac{dy}{dx}$ or $\frac{dx}{dy}$. Please edit the question accordingly. – Srivatsan Oct 24 '11 at 23:54
It turns out I waited long enough. The answer in your comment is right. – André Nicolas Oct 24 '11 at 23:57
If you are finding $\frac{dy}{dx}$ then you are correct; it is $\frac{-18x}{-2y}$. – Joe Johnson 126 Oct 24 '11 at 23:59
Some of us would then simplify to $(9x)/y$. – Gerry Myerson Oct 25 '11 at 0:11
up vote 2 down vote accepted

$$ \frac{d}{dx}\left(9x^2 - y^2\right) = 18x - 2y\cdot\frac{dy}{dx} = \frac{d}{dx} 1 = 0. $$ Then from $$ 18x - 2y\cdot\frac{dy}{dx} = 0, $$ you can find $dy/dx$.

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(+1), also to get the ball rolling ;) – The Chaz 2.0 Oct 25 '11 at 2:15

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