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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. – Henning Makholm Dec 12 '11 at 23:06
up vote 3 down vote accepted

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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Thanks for the response. Is there a particular rule that you used to derive the second part? Is it the chain rule? – lord12 Dec 12 '11 at 23:49
Yes, it's the chain rule. $df/du=(df/dy)(dy/du)$ – Gerry Myerson Dec 13 '11 at 0:34

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