I had this question on a quiz and I answered 35!x^5/4!
What is the 35th derivative of $f(x) = e^{x^{10}} $at $x = 0$? Use a suitable Taylor Polynomial for $e^x$ at $x = 0$. Express the answer in terms of factorials.
I substituted $x^{10}$ into the standard taylor polynomial for $e^x$, so I got
$$1+x^{10}+x^{20}/2!+x^{30}/3!+x^{40}/4!$$
Now that I think about it, it isn't quite 35!, because there is still 5!, after the 35th derivative. So I should have:
$$\frac{35!}{5!} \frac{x^5}{4!}$$, is this correct?
But at $x = 0$, would the answer be $0$?
Thanks for any feedback.