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There are many well-known functions (such as $x^x$) which have no closed-form integral. However, are there any elementary functions whose derivatives cannot be expressed in a closed-form manner?

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Is $x^x$ an elementary function? –  Hui Yu Dec 20 '12 at 1:36
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All elementary functions have derivatives. Compositions and arithmetic operations have rules for derivatives. Consequently all elementary functions have closed-form expressions for derivatives. –  copper.hat Dec 20 '12 at 1:38
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1 Answer

up vote 12 down vote accepted

If $f(x)$ is built out of the "usual" functions (rational functions, exponentials, logarithms, trig, inverse trig) using addition, subtraction, multiplication, division, or composition, then $f'(x)$ is also.

The proof is on induction on complexity. For the base case, all of "easy" functions work: If $f(x) = x$ or $e^x$ or $\ln x$ or $\sin x$ or $\operatorname{trig}(x)$ or $\operatorname{arctrig}(x)$ (where "trig" stands for one of the 6 trig functions), then $f'(x)$ is elementary.

Next, suppose $f(x)$, $f'(x)$, $g(x)$, and $g'(x)$ are all elementary.

Is $f(x) + g(x)$ also elementary? Yes because $(f(x)+g(x))' = f'(x) + g'(x)$ is a sum of elementary functions, hence is elementary. The same argument works for subtraction.

For multiplication and division, use the product/quotient rules. For composition, use the chain rule.

Then, e.g., composition works as follow: $f(g(x))' = f'(g(x))g'(x)$. But $f'(g(x))$ is elementary because $f'$ and $g$ are and elementary things are closed under composition. Further, $g'(x)$ is elementary by assumption. So, $f'(g(x))g'(x)$ is a product of elementary functions, hence elementary.

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