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Mathematics
Calculus
Does derivative formula work for all parametric equations
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[QUOTE="fresh_42, post: 5503665, member: 572553"] You could resolve it by separately consider the two branches ##y = ± \sqrt{1-x^2}##. Whether you can explicitly write ##y=y(x)## or not doesn't change the dependency. It makes the difference between ##\frac{dy}{dx} = 0## and ##\frac{dy}{dx} ≠ 0.## Your example becomes $$0 = \frac{d}{dx} 1 = \frac{d{x^2}}{d{x}} + \frac{d{y^2}}{d{x}} = 2x + 2y \frac{d{y}}{d{x}} \text{ and therefore } x+ y \cdot \frac{d{y}}{d{x}} = 0$$ If ##y## and ##x## were independent then ##x## would be a constant function and not a coordinate of a circle. [/QUOTE]
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Mathematics
Calculus
Does derivative formula work for all parametric equations
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