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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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One unfortunate artifact of this notation: I've seen more than one person make the mistake of thinking that the $dx^2$ in the denominator represents $d(x^2)$ rather than $(dx)^2$... –  Steven Stadnicki May 31 '13 at 22:10

4 Answers 4

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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That second example is a bit unfortunate, since $[g(x)]^2 = (x^2)^2$ as well. But I disagree that $g^2(x)$ unambiguously means $(g \circ g)(x)$, as $\sin^2(x)$ certainly does not mean $\sin(\sin x)$. –  TMM May 31 '13 at 15:26
    
I meant that the exponent in the case of the second derivative stands for composition. You are correct that it is not unambiguous in mathematics. I will edit the example. –  Kris Williams May 31 '13 at 15:54

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.

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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.

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