derivative logical meaning basic question

Question

The question is about the logical meaning of $\frac{df(x-1)}{dx}$, I did a single transform, which is $\frac{df(x-1)}{dx} = \frac{df(x-1)}{d(x-1)} * \frac{d(x-1)}{dx}$, since latter part $\frac{d(x-1)}{dx}$ is $1$, so $\frac{df(x-1)}{dx} = \frac{df(x-1)}{d(x-1)} = \frac{df(x)}{d(x)}$?

To me it seems $\frac{df(x-1)}{dx}$ is not the same as $\frac{df(x)}{d(x)}$? If anyone could explain the logical meaning of $\frac{df(x-1)}{dx}$ and how it equals to $\frac{df(x)}{d(x)}$, it will be great. :)

BTW, similarly, I can also get $\frac{df(1-x)}{dx}$ is $-\frac{df(x)}{dx}$

Please any steps is not correct, please feel free to correct me.

thanks in advance, Lin

Answer 1

Let’s look at the chain rule using a different notation, since I suspect that’s what’s tripping you up. It says that $$(f\circ u)'(a) = f'(u(a))\,u'(a).$$ Note that the derivative of $f$ is taken at $u(a)$. This is an important detail that in my opinion the Leibnitz notation obscures. (You can find a similar, but probably better-presented, explanation of the chain rule in this answer.)

In your case, we have $u:\xi\mapsto \xi+1$, so $(f\circ u)'(x) = f'(x+1)\cdot1 = f'(x+1)$. This makes intuitive sense: $f\circ u$ is just $f$ shifted to the left by one unit, so one would expect $(f\circ u)'$ to be $f'$ shifted left by one as well.

Using Leibnitz notation, you started off well enough, but appear to have lost the fact that the derivative of $f$ must be taken at $x+1$. Keeping explicit track of where the derivatives are evaluated, we have $${df(\xi+1)\over d\xi}\Big|_{\xi=x} = {df(\xi+1)\over d(\xi+1)}\Big|_{\xi=x}{d(\xi+1)\over d\xi}\Big|_{\xi=x} = {df(\eta)\over d\eta}\Big|_{\eta=x+1}\cdot1\Big|_{\xi=1} = {df(\eta)\over d\eta}\Big|_{\eta=x+1}.$$ I’ve introduced some new variables to distinguish what are effectively dummy variables from the points at which the derivatives are evaluated. This is more or less te derivation you had, except that derivative of $f$ is evaluated at $x+1$ instead of at $x$. A similar derivation gives $${df(1-\xi)\over d\xi}\Big|_{\xi=x} = -{df(\eta)\over d\eta}\Big|_{\eta=1-x}$$ which is again what you had except evaluated at $1-x$.

As a concrete example, take $u:(\cdot)\mapsto(\cdot)+1$ again and let $f:(\cdot)\mapsto (\cdot)^3-2(\cdot)+1$. (Here I’m using $(\cdot)$ to stand for the function parameter so that there’s no chance of confusing it with the variables in the arguments to the functions.) The derivative of $f$ at any point is given by $f':(\cdot)\mapsto 3(\cdot)^2-2$ and the chain rule essentially tells us that we can find $(f\circ u)'(x)$ by plugging $u(x)=x+1$ into $f'$ (and multiplying by $u'(x)$, which is $1$ here). Doing so gives us $3(x+1)^2-2 = 3x^2+6x+1$. We check this by expanding $f(x+1)$ and differentiating the result: $$f(x+1)=(x+1)^3-2(x+1)+1=x^3+3x^2+x$$ and differentiating this yields $3x^2+6x+1$ again.