Why can't one cancel partial differentials?

Question

I have a question about the below formula:

$$\frac{dz}{ds} = \frac{dz}{dx} \cdot \frac{dx}{ds} + \frac{dz}{dy} \cdot \frac{dy}{ds}$$

Ok. I understand what this means. Small change of s makes small change of x and y and thus the sum makes small change of z.

However, I simply can't understand the formula itself.

$$2\cdot\frac{dz}{ds} = \frac{dz}{dx} \cdot \frac{dx}{ds} + \frac{dz}{dy} \cdot \frac{dy}{ds}$$

Why not this? Every book I have read defines calculus by multiplication and division. Then it must be $2\cdot\frac{dz}{ds}$ I think. Is there anything I missed?

edit: z= z(x,y), x=x(s,t), y=y(s,t)

Answer 1

Writing the formula like this is wrong for $z\equiv z(x,y), x\equiv x(s,t), y\equiv y(s,t)$

One should write it using partial derivatives. \begin{align*} dz &= \frac{\partial z}{\partial x}dx+ \frac{\partial z}{\partial y}dy\\ dz &= \frac{\partial z}{\partial x}\left(\frac{\partial x}{\partial s}ds+\frac{\partial x}{\partial t}dt \right)+ \frac{\partial z}{\partial y}\left(\frac{\partial y}{\partial s}ds+\frac{\partial y}{\partial t}dt\right)\tag{1}\\ \frac{\partial z}{\partial s} &= \left(\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\right)ds+\left(\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}\right)dt\\ \frac{\partial z}{\partial s} &= \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\tag{2}\\ \end{align*} If it were $z\equiv z(x,y), x\equiv x(s), y\equiv y(s)$, then it would have become \begin{align*} dz &= \frac{\partial z}{\partial x}\frac{dx}{ds}ds+ \frac{\partial z}{\partial y}\frac{dy}{ds}ds\\ \frac{dz}{ds} &= \frac{\partial z}{\partial x}\frac{dx}{ds}+ \frac{\partial z}{\partial y}\frac{dy}{ds}\\ \end{align*} If it were $z\equiv z(x), x\equiv x(s,t)$, then it would have become \begin{align*} dz &= \frac{dz}{dx}\left(\frac{\partial x}{\partial s}ds+\frac{\partial x}{\partial t}dt \right)\\ \frac{\partial z}{\partial s} &= \frac{dz}{d x}\frac{\partial x}{\partial s}\\ \end{align*} If it were $z\equiv z(x), x\equiv x(t)$, then it would have become \begin{align*} dz &= \frac{dz}{dx}\frac{dx}{dt}dt\\ \frac{dz}{dt} &= \frac{dz}{d x}\frac{dx}{dt}\tag{3}\\ \end{align*} This is why one can write $a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$ in a one-dimensional mechanics problem. Here, $dx$ seem to cancel but what is going on is equation $(3)$.

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This apparent cancellation doesn't take place on partial differential $\partial x$ as in equation $(2)$. For example: In polar coordinates, $x\equiv x(r,\theta)=r\cos\theta,y\equiv y(r,\theta)=r\sin\theta,r\equiv r(x,y)=\sqrt{x^2+y^2},\theta\equiv\theta(x,y)=\tan^{-1}\frac yx$

• $\frac{\partial x}{\partial r}=\cos\theta$ if one takes $r,\theta$ as independent variables.
• $\frac{\partial r}{\partial x}=\cos\theta$ if one takes $x,y$ as independent variables.

Clearly, $\frac{\partial x}{\partial r}\frac{\partial r}{\partial x}\neq1$ generally.

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Takeaway: Always start with equations like equation $(1)$ taking into account all the dependencies of all the variables.