Why is the derivative of this Gaussian Function negative before being positive?

Question

From pg. 61 of Principles of Quantum Mechanics, the Guassian $g_Δ$ is defined as

$$ g_Δ(x-x') = \frac{1}{(πΔ^2)^{1/2}} \exp \left[ -\frac{(x-x')^2}{Δ^2} \right] $$

where the Gaussian is centered at $x'=x$, has width $\Delta$, maximum height $(π Δ^2)^{-1/2}$, and has unit area (independent of $Δ$). In case it matters, we can assume that $Δ ∈ \mathbb{R}$.

Problem: The book goes on to display the derivative of $g_\Delta$ as follows:

It seems to me the derivative graph (b) is wrong, since it is negative when $x' < x$ and positive when $x' > x$. Isn't this exactly backwards, and if not, why?

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EDIT for bounty: The crux of my confusion lies in understanding/getting a different viewpoint for why

$$ \frac{dg_\Delta(x-x')}{dx} = -\frac{dg_\Delta(x-x')}{dx'} $$

Moreover, is this actually a general theorem, or is this using something special about $g_\Delta$ in particular? That is, if we let $x,' \in \mathbb{R}$ and consider $f(x-x')$ as an arbitrary real function, do we have in general that

$$ \frac{df(x-x')}{dx} = -\frac{df(x-x')}{dx'}? $$

Answer 1

This is due to the chain rule.

For any differentiable function,

$$\frac d{dx}f(x-x')=-\frac d{dx'}f(x-x')$$ just because

$$\frac d{dx}(x-x')=1$$ and $$\frac d{dx'}(x-x')=-1.$$

The plot in the book is correct because the dérivative is on $x$, while the axis is $x'$.

Answer 2

Rotate your coordinate system (x,x') by 45 degrees. For a function $g_\Delta (x-x')$ even in $x-x'$, you have $$ \frac{ d g_\Delta (x-x')}{dx} =- \frac{d g_\Delta (x-x')}{d x' }, \tag{1.10.20} $$ so the ordinate of (b) is an odd function, by convention called $ g_\Delta ' (x-x')$, but really $- d g_\Delta (x-x')/d x' $, so minus the derivative of (a) w.r.t. x'.

To the left of x, this is negative, because $ g_\Delta ' (x-x')>0.$ Its minimum is at $x'=x-\epsilon$, and its maximum is at $x'=x+\epsilon$. So $$ g_\Delta ' (x-x') \sim \delta (x'-\epsilon-x) -\delta(x'-x +\epsilon ) $$ sampling the function f as the author claims, correctly.

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Note on your additional edit:

Define $y\equiv x-x'$, the light-cone variable. It then follows that $$ \frac{d }{dx} f(y) + \frac{d }{dx'} f(y) = \frac{d y }{dx} \frac{d f(y) }{dy} + \frac{d y }{dx'} \frac{d f(y) }{dy}\\ = \frac{d f(y) }{dy} - \frac{d f(y) }{dy}=0. $$

Answer 3

Let $$ h(x, x') = \frac{1}{(\pi \Delta^2)^{1/2}}e^{-\frac{(x - x')^2}{\Delta^2}}. $$ Notice that $$ \frac{\partial h(x,x')}{\partial x} = \frac{1}{(\pi \Delta^2)^{1/2}}e^{-\frac{(x - x')^2}{\Delta^2}} \left(-\frac{2(x - x')}{\Delta^2} \right). $$ Plotted as a function of $x$, with $x'$ held fixed, the graph of $\frac{\partial h(x,x')}{\partial x}$ looks like this, as you would expect:

                  

However, the author Shankar (confusingly) plots $\frac{\partial h(x,x')}{\partial x}$ as a function of $x'$, with $x$ held fixed. The resulting plot looks like this:

                  

Answer 4

Perhaps this rewriting is clear. The crux of the issue seems to be that $$ \xi(x,x'):=x-x'$$ satisfies $$ \partial_x\xi(x,x')=1=-\partial_{x'}\xi(x,x').$$ Here $\partial_x := \frac{d}{d x}$ is shorthand. Then for each fixed $x,x'$, and any one variable function $f$, we can apply chain rule $$ \partial_x (f\circ \xi)(x,x') = f'(\xi(x,x'))\partial_x \xi(x,x') = - f'(\xi(x,x'))\partial_{x'} \xi(x,x') = - \partial_{x'} (f\circ \xi)(x,x').$$ This is what I would understand as the meaning of $$ \frac{df(x-x')}{dx} = -\frac{df(x-x')}{dx'}.$$ One could also proceed without chain rule; since $$ \frac{f((x+h)-x')-f(x-x')}{h} = \frac{f(x-(x'-h))-f(x-x')}{h} = -\frac{f(x-(x'-h))-f(x-x')}{-h}$$ we have \begin{align} \frac{df(x-x')}{dx} &=\lim_{h\to0} \frac{f((x+h)-x')-f(x-x')}{h}\\ &=-\lim_{h\to0} \frac{f(x-(x'-h))-f(x-x')}{-h}\\ &=-\lim_{h\to0} \frac{f(x-(x'+h))-f(x-x')}{h}\\ &=-\frac{df(x-x')}{dx'}. \end{align}