Calculating $g^{(1000)}(0)$ of $g(x)=\exp(x^{999})$

Question

$g^{(1000)}(0)$ of $$g(x)=\exp(x^{999})$$

I know that we can write $$g(x)=\exp(x^{999}) = \sum_{n=0} \frac{x^{999n}}{n!}$$

So I looks like it should be $0$ but what's the reasoning behind it?

Answer 1

Hint: first compute $g'(x)$. You get a product of two terms. Now take $g''(x)$ using the product and chain rules. You should get something that looks like $p(x)\exp(x^{999})$ where $p(x)$ is a polynomial. You should be able to imagine what will happen with $998$ more derivatives.

Answer 2

Your way is also fine: $$g(x)=e^{x^{999}}=1+x^{999}+\frac{x^{1998}}{2}+\frac{x^{2997}}{6}+\cdots.$$ $$g^{(1000)}(x)=\frac{1998\cdots 999 x^{998}}{2}+\frac{2997\cdots 1998x^{1997}}{6}+\cdots.$$ $$g^{(1000)}(0)=0.$$