im asked to find the derivative of this function using the chain rule.

here are my steps.

step 1 - identify the inner and outer functions. therefore I identified outer function as $e^x$ inner function as $3\sqrt{x}$

step 2- i used derivative of outer function with respect to inner times the derivative of inner function. so the answer as i see it should be $$e(3\sqrt{x})\left(\frac{3}{2}x^{-1/2}\right) $$

however the answer is $$e^{3\sqrt{x}}\left(\frac{3}{2}x^{-1/2}\right)$$

what I'm missing? I know the derivative of $e^x = e^x$ is that were i'm going wrong? Thanks in advance for any explanation clarification you guys can offer. Miguel

  • 3
    $\begingroup$ The chain rule states $(e^u)' = e^u u'$. You have written $(e^u)' = e(u)u'$. $\endgroup$
    – JavaMan
    Mar 2, 2013 at 3:31
  • $\begingroup$ thanks javaman so those that mean i treat the $e^x$ differently if i'm understanding you correctly? thats the reason is just itself. $\endgroup$
    – Miguel
    Mar 2, 2013 at 3:32
  • $\begingroup$ $\frac{3 e^{3 \sqrt{x}}}{2 \sqrt{x}}$ $\endgroup$
    – Arjihad
    Jun 30, 2018 at 15:52

3 Answers 3


For the exponential function, $$\frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x).$$

Here, $e^{f(x)} = e^{3\sqrt x}$, so $f(x) = 3\sqrt x = 3x^{1/2}$. So then we must have $$\frac{d}{dx}\left(e^{3\sqrt{x}}\right) = e^{3\sqrt{x}}\left(\frac{3}{2}x^{-1/2}\right) = \frac{3e^{3\sqrt x}}{2\sqrt x}$$

As per comments: Yes, $e^x$ is unique in comparison to $x^n$, in many ways, including the fact that the first is a very distinguished constant raised to a variable power whereas the second is a variable raised to a constant. So the power rule does not apply to $e^x$, nor does it apply to any constant raised to a variable. And with the unique constant $e$: recall, $\;\frac{d}{dx}(e^x) = e^x$.

  • $\begingroup$ thank you amWhy i see the reason why my answer is not correct now. $e$ is unique in comparison to say $x^u$. i see it now. I appreciate your response. $\endgroup$
    – Miguel
    Mar 2, 2013 at 3:38
  • 1
    $\begingroup$ You're welcome. Yes, we DO NOT use the power rule with $e^x$, like we might with $x^n$, for a number of reasons. To your credit: You identified the correct inner and outer functions! $\endgroup$
    – amWhy
    Mar 2, 2013 at 3:41
  • $\begingroup$ did it nicely :+) $\endgroup$
    – Mikasa
    Mar 2, 2013 at 5:03

Note that $\displaystyle \frac d {dx}e^x=e^x$, not $ex$.


A slightly different way of doing it by logging both sides (here $L(\cdot)=\log(\cdot))$: $$ f(x) =e^{3 \sqrt{x}}\\ L(f(x))=3 \sqrt{x}\\ \frac{f'(x)}{f(x)}=\frac{3}{2 \sqrt{x}}\\ f'(x)=f(x)\frac{3}{2 \sqrt{x}}= \frac{3}{2 \sqrt{x}} e^{3 \sqrt{x}} $$ This is due to $(\log f(x))'_{x}=\frac{f'(x)}{f(x)}$


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