$$f(x) = \frac{x}{e^{x^2}}$$ Differentiate $f(x)$.

How should the above function be interpreted? Is the function equivalent to:

a)$$f(x) = \frac{x}{e^{x^2}} = \frac{x}{{(e^x)^2}} = \frac{x}{e^{2x}}$$


b) $$f(x) = \frac{x}{e^{x^2}} = \frac{x}{{e^{(x^2)}}} ≠ \frac{x}{e^{2x}}$$


  • 7
    $\begingroup$ It is interpreted as (b). $\endgroup$
    – LASV
    Dec 13, 2013 at 5:09
  • $\begingroup$ @dfg:Maybe helps to use some small, simple values like $x=1$. $\endgroup$ Dec 13, 2013 at 5:16

2 Answers 2


$a^{b^c}$ should be interpreted as $a^{(b^c)}$. Because, as you point out, $(a^b)^c = a^{bc}$. This is even true in most programming languages, which specify that a^b^c should be evaluated from right to left, unlike expressions like a-b+c, which should be evaluated from left to right.

  • $\begingroup$ Out of curiosity, what programming languages use ^ for exponentiation? $\endgroup$ Dec 13, 2013 at 5:14
  • $\begingroup$ BASIC, MATLAB, and Mathematica are examples. $\endgroup$ Dec 13, 2013 at 5:16
  • 1
    $\begingroup$ Ah. I use those so infrequently I sometimes forget there are languages where it isn't bitwise XOR. $\endgroup$ Dec 13, 2013 at 5:27
  • $\begingroup$ But I just tried out 2^3^4 in OCTAVE, which is meant to emulate MATLAB, and it did the computation left to right. But Mathematica did it right to left. I am pretty sure that BASIC does it right to left, because I remember reading about it in a BASIC manual, and wondering why. $\endgroup$ Dec 13, 2013 at 5:27
  • $\begingroup$ FORTRAN uses **. It does it right to left: folk.uio.no/hpl/scripting/doc/f77/tutorial/expres.html $\endgroup$ Dec 13, 2013 at 5:28

The expression

$f(x) = x / e^{x^2} \tag{1}$

should be interpreted in the same way we would interpret $x / exp(x^2)$;that is,

$f(x) = x e^{-x^2} = x \exp (-x^2). \tag{2}$

The tip-off here is that the exponent $2$ of $x$ in the expression $x^2$ is in fact written as a superscript of $x$; in item (a) of the question, $(e^x)^2 = (e^x)(e^x) = e^{2x}$; but $e^{x^2}$ should be thought of as $e^{xx} = (e^x)^x$; as Stephen Montgomery points out in his answer, the evaluation of the exponents is right to left.

Having hopefully clarified the issue, we have

$f'(x) = (x e^{-x^2})' = e^{-x^2} + x e^{-x^2} (-2x) \tag{3}$

using the Leibniz product rule and the formula $d / dx (e^u) = e^u (du / dx)$. After the simple algebra,

$f'(x) = e^{-x^2}(1 - 2x^2). \tag{4}$

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.