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I just read that:

If $z=f(x,y)=c$, be the equation of a curve, then the slope of the tangent to the curve at any point (x,y), is given by $$m=\frac {dy}{dx}=-\frac{\frac{\partial z}{\partial x}}{\frac {\partial z}{\partial y}}$$

I don't see how the minus sign creeps in here.(Of course I don't have a proof, but the - sign is against intuition).

A proof(or a link to a simple proof) would be nice, and an intuitive explanation would be nicer. Thanks for help.

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    $\begingroup$ Example 1, example 2 and more generally see the statement of the implicit function theorem given here. Note the part that says "Furthermore, $J_G=-\left(J_2\right)^{-1}J_1$". $\endgroup$ – Git Gud May 23 '14 at 13:53
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As $z=f(x,y)=c$ $$\bigtriangledown z=\frac{\partial z}{\partial x}dx+\frac {\partial z}{\partial y}dy=0$$ so$$ \frac {dy}{dx}=-\frac{\frac{\partial z}{\partial x}}{\frac {\partial z}{\partial y}} $$

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    $\begingroup$ After doing a bit of reading on the internet, I (think I) finally understood your solution. What I learned besides was that I only knew 1% of the chain rule. $\endgroup$ – Shubham May 23 '14 at 14:15
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This can be intuitively understood by realizing that we are calculating the Slope of the Tangents to the Level Curves of a Surface.

Level Curves (and thus their Tangents) always run Perpendicular to the Gradient Vector of the Surface.

The Gradient Vector has Slope ${f_y\over f_x}$.

The Slope of a Line Perpendicular to another has as Slope the Negative Reciprocal of the other Line's Slope.

Therefore the Implicit Derivative is $-{f_x\over f_y}$ because this is the Negative Reciprocal of the Slope of the Surface $f(x,y)$'s Gradient.

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