Is it possible to take a second derivative without taking the first derivative before?
Why do we multiply the $d$ and $dx$ operators? Like, does $\dfrac{d^2}{dx^2}$ really mean $\dfrac{d}{dx} \cdot \dfrac{d}{dx}$?
What's the intuitive understanding about this? Can it be represented in a graph? Like... 'Little change squared in $y$ over little change squared in $x$'?
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I'd prefer to view $\frac{d^2}{dx^2}$ as $\frac{d}{dx}\circ\frac{d}{dx}$. You can hardly avoid taking the first derivative before computing the second. You might define $f''(x_0)$ as the unique number $a$ such that $f(x)-a(x-x_0)^2$ is of the form $y_0+mx+o((x-x_0)^2)$, but in that case you "accidentally" have obtained the first derivative $m$.
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Even though the notation looks like multiplication, it is really function composition. That is $$ \frac{d^2}{dx^2} \left[ f(x) \right] = \frac{d}{dx} \left[ \frac{d}{dx} \left[f(x)\right]\right].$$
In terms of simpler operations, consider the function $g(x)=x^4$ where $x$ is a real number. Then $$g^2(x) = (g \circ g)(x) = g(g(x)) = g(x^4) = (x^4)^4.$$ The exponent is a shorthand for composition, not multiplication.
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If you rather think in terms of rise and run and the limit through $f' \approx {{\Delta y}\over {\Delta x}}= {{f(x+h)-f(x)}\over{h}}$ then you can define the second derivative through $f'' \approx {{f(x+h)-2f(x)+f(x-h)}\over{h^2}}$. Now the numerator is the difference of two, left and right, rises as in $f(x+h)-2f(x)+f(x-h)=[f(x+h)-f(x)]-[f(x)-f(x-h)]$.
So in some sense this is a rise over run of a rise over run.
It depends what you mean by take a second derivative without taking the first derivative before. You can use formulae to calculate a second derivative without calculating the first derivative but this formula would have relied on first taking the first derivative.
The second derivative is the little change in the first derivative over the little change in x.