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The physical meaning of the differentiation of $x$ with respect to $y$ is the rate of change of $x$ with respect to $y$. But, I am finding it difficult in understanding the geometrical interpretation of the expression $\dfrac{\mathrm{d}x}{\mathrm{d}y}$. On a given curve on any given points, what does the expression actually tells us?

I currently know that differentiating the equation of a curve gives the tangent of that curve. But how can we understand that tangent line in terms of $y$ and $x$?

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  • $\begingroup$ Differentiating the equation of a curve does not give you the tangent line, it gives the slope of the tangent line $\endgroup$ – GFauxPas Jun 28 '15 at 14:33
  • $\begingroup$ You can consider that $x$ depends on $y$, that is $x(y)$. $\endgroup$ – UnknownW Jun 29 '15 at 7:06
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Differentiating $x$ with respect to $y$ gives you the gradient of the tangent at the point $\left(x, y(x)\right)$ on the curve $y(x)$. The gradient of the tangent at the point $\left(x, y(x)\right)$ indicates the rate of change of $y(x)$ at that specific point. Because on a curve, the gradient varies, differentiating is basically taking the limit of the $\dfrac{\Delta y}{\Delta x}$ as $\Delta x \to 0$ and hence finding the gradient or rate of change of a function at a specific point on a curve.

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Remember $y$ is just a shorthand notation for $y(x)$ as in $y$ is a function that takes a number as an input and outputs a number. Think of $\frac{d}{dx}$ as a function that takes a function as input and outputs a function. Don't forget the definition $ \frac{d}{dx}y(x) = \lim_{h->0} \frac{y(x+h)-y(x)}{h}$. So the outputted function gives us information about how the function acts on points close to $x$.

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  • $\begingroup$ The OP says $dx/dy$, not $dy/dx$. $\endgroup$ – UnknownW Jun 29 '15 at 7:03

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