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How do you differentiate $\frac{d}{dx^2}f(x)$, where $f(x) = x$?

Attempt: I know $\frac{d}{dx}f(x) = 1$, but I am not sure what to do when you differentiate with respect to a higher order term.

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    Are you sure it's not $\frac{d^2}{dx^2}f(x)$? That means the derivative of the derivative, which is $0$ in this case.2011-12-12

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Do you really mean ${d\over dx^2}f(x)$? or do you mean ${d^2\over dx^2}f(x)$?

If the latter, it's just the derivative of the derivative (which, in your example $f(x)=x$, is just $0$).

If the former, I guess ${d\over dx^2}f(x)={df(x)\over dx}\cdot{dx\over dx^2}={df(x)\over dx}\div{dx^2\over dx}={df(x)\over dx}\div(2x)={1\over2x}$ in your case.

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    Yes, it's the chain rule. $df/du=(df/dy)(dy/du)$2011-12-13