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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.

  • 2
    The $35$-th *derivative* is a certain infinite sum whose leading term is $\frac{40!}{5!4!}x^5$. But there is a next term, and one after that, forever. (At $0$ it is $0$.)2012-06-02

2 Answers 2

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The coefficient for $x^{35}$ in the Taylor expansion for $x\mapsto \exp\{x^{10}\}$ is 0. Hence its 35th derivative at 0 is 0.

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Hint Is $35$ a multiple of $10$?

Remember that given $f(x)$, the coefficient of $(x-a)^n$ in the Taylor expansion around $x=a$ is precisely $f^{(n)}(a)/n!$