When $ y = v + 1 $ and $ x = u + 3 $, is it true that $ dy/dx = dv/du $?

Question

Let $ y(x) $ be a function of $ x $.

Let $ u(x) = x - 3 $.

Let $ v(y) = y - 1 $.

Is it true that $ \frac{dy}{dx} = \frac{dv}{du} $?

If yes, how can we prove it?

If not, what is a simple counterexample for it?

Answer 1

It is indeed true that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}v}{\mathrm{d}u}$. Notice that $\frac{\mathrm{d}v}{\mathrm{d}u}=\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}u}=\frac{\mathrm{d}v}{\mathrm{d}x}$. Furthermore, $y=v+1$ implies $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}v}{\mathrm{d}x}$. Thus $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}v}{\mathrm{d}x}=\frac{\mathrm{d}v}{\mathrm{d}u}$$

Answer 2

Sometimes the Leibniz notation $\frac{dy}{dx}$ is confusing. [In particular using the letter $y$ as both a function and a variable at the same time can confuse beginners.] Let's analyze the meaning.

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Writing $\frac{dy}{dx}$, you assume $y$ is a function of $x$. Say $y = \phi(x)$, where $\phi$ is a function. Then $$ \frac{dy}{dx} = \phi'(x) . $$
Next, in order to write $\frac{dv}{du}$, we need to write $v$ as a function of $u$. We are given $u=x-3, v=y-1$. Then we write $v$ as a function of $u$ by computing: $$ v = y-1 = \phi(x)-1 = \phi(u+3)-1 . $$ So $$ \frac{dv}{du} = \frac{d}{du}\big(\phi(u+3)-1\big) = \phi'(u+3) $$ We used the chain rule, together with the observation that both $3$ and $-1$ have derivative $0$. Finally, in terms of $x$ $$ \frac{dv}{du} = \phi'(u+3) = \phi'(x) = \frac{dy}{dx} . $$

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Note: Here I used the letter $x,y,u,v$ as variables, and the letter $\phi$ as a function. I wrote $y = \phi(x)$ and did not write confusing things like $y(x), u(x), v(y), v(u)$.

Answer 3

It's true

${y}={v}+\mathrm{1}\:,\:{x}={u}+\mathrm{3} \\ $ $\frac{{dy}}{{dx}}=\frac{{d}\left({v}+\mathrm{1}\right)}{{dv}}.\frac{{dv}}{{du}}.\left(\frac{{d}\left({u}+\mathrm{3}\right)}{{du}}\right)^{-\mathrm{1}} =\frac{{dv}}{{du}} \\ $