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In maths class i have been taught not to treat ${dy\over dx}$ as a ratio but in physics why do we treat it like one.

$$dw = f ds \implies dw = m \times {dv\over \color{red}{dt}} \times ds \implies dw = m\times dv\times{ds \over \color{red}{dt}}$$

$$\text{w is work done, f is force, v is velocity, s is displacement, t is time and m is mass}$$

If ${dy\over dx}$ is not a ratio then how did it change its numerator ? I think it will be clear if somebody can rewrite the above equation in Lagrange's notation. I hope i did not trouble anyone by asking this question. Thanks.

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  • $\begingroup$ it happens to work, the reason is, I believe, explained in analysis courses. $\endgroup$ – Alvin Lepik Sep 7 '16 at 9:07
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    $\begingroup$ @AlvinLepik I am not interested in working; rather why it is working is my concern ? please explain me. $\endgroup$ – user366914 Sep 7 '16 at 9:11
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    $\begingroup$ Roughly speaking $\frac{dy}{dx}$ is not a ratio in a restricted domain of mathematics. But a larger view is provided by Non-Standard Analysis. For example, see : en.wikipedia.org/wiki/Non-standard_analysis . This makes perfectly consistent the symbolism $\frac{dy}{dx}$ with the traditional use of it in physics. So, don't worry, you can use $\frac{dy}{dx}$ as a ratio in current applications. In more special cases (certainly not in physics, but in very specific field of mathematics), one have to mention some criticisms about Non-Standard Analysis. $\endgroup$ – JJacquelin Sep 7 '16 at 9:32
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    $\begingroup$ This is not nearly as mysterious as one might think. You can just think of $dy$, $dx$, etc as extremely tiny but finite real numbers. Then your equations only hold approximately, but the approximations are very, very good, and you can hope that "in the limit" you will obtain exact equality (if you're careful). $\endgroup$ – littleO Sep 7 '16 at 9:35
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    $\begingroup$ The question $\frac{dy}{dx}$ is or isn't a ratio has long been a debate among Mathematicians and Physicians. Nowadays many of them are considering that the theoretical aspect of the question is overcome by Non-Standard Analysis. But indeed for a long time before that, on practical viewpoint considering $\frac{dy}{dx}$ as a ratio was accepted in Physics and in low-level Mathematics as well. For French readers, a paper for the general public : fr.scribd.com/document/14755203/… $\endgroup$ – JJacquelin Sep 7 '16 at 10:24
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This is one of the reasons that Leibniz notation is often described as 'powerful'. Using the Calculus of Limits,

$\frac{dv}{dt}ds =\lim_{\delta t \to 0}{\frac{v(t+\delta t) - v(t)}{\delta t}} . [s(t+\delta t) - s(t)] = \lim_{\delta t \to 0}v(t+\delta t) - v(t) . \frac{s(t+\delta t) - s(t)}{\delta t}=dv\frac{ds}{dt}$

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    $\begingroup$ Why are $\lim_{\delta t \to 0}{\frac{v(t+\delta t) - v(t)}{\delta t}} . [s(t+\delta t) - s(t)]$ or $\lim_{\delta t \to 0} [v(t+\delta t) - v(t)] . \frac{s(t+\delta t) - s(t)}{\delta t}$ not zero? $\endgroup$ – Henry Sep 7 '16 at 12:00
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I would regard $dw = f \, ds$ as an abuse of notation for either $f=\dfrac{dw}{ds}$ or $\displaystyle \int dw = \int f \, ds$.

That being said, in analysis you can show the chain rule: $\dfrac{dz}{dx}= \dfrac{dz}{dy}\dfrac{dy}{dx}$ and the inverse function rule $\dfrac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}}$ under specified conditions, and these will lead to what you call commutative multiplication

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  • $\begingroup$ Very good answer. +1 The actual expression here should be $f = dw/ds$ and this can be used to prove that $dw/dv = mv$ where $v = ds/dt$ and $f = mdv/dt$ $\endgroup$ – Paramanand Singh Sep 7 '16 at 9:32
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The differentials $dy$ and $dx$ are new variables which are related to each other by $dy = f^{\prime}(x) \; dx$, which is the same as $dy = \frac{dy}{dx} \; dx$. One way to think of the differentials is to pick a point on the curve $y=f(x)$ and make that point the origin of a $dx-dy$ coordinate system, where the $dx$-axis is parallel to the $x$-axis and m.m. $dy$ and $y$. (It's sort of a traveling co-ordinate systems that moves along with your choice of $x$.) The tangent line to the curve at that point has equation $dy = f^{\prime}(x) \; dx$ in the $dx-dy$-plane. Because of the equation of this line, we can replace $dy$ by $\frac{dy}{dx} \; dx$. In your example, we take $$m \times \frac{dv}{dt} \times ds$$ and multiply through by $dt$ to get $$m \times \frac{dv}{dt}\times dt \times ds.$$ Then we replace the $\frac{dv}{dt}\times dt$ with $dv$ to get $$d \times dv \times ds.$$ Now we have to pay for multiplying by $dt$, so we replace $ds$ bye $\frac{ds}{dt} \times dt$ to get $$m\times dv \times $\frac{ds}{dt} \times dt.$$ Then we divide by $dt$ to make up for multiplying by $dt$ earlier and get $$m\times dv \times \frac{ds}{dt}.$$ All these "replacements" look exactly like cancelling (or un-cancelling) a $dt$.

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