100th derivative of $ g(x) = (x^3 -4x + 7)^{30} $. Where is the pattern?

Question

I was reviewing come Calc 1 problems, and came across one that asks us to explain what the 100th derivative of

$$ g(x) = (x^3 -4x + 7)^{30} $$

would be. I computed the first three derivatives:

$$ g'(x) = 30(x^3-4x+7)^{29}(3x^2-4) $$ $$ g''(x) = 30*29(x^3-4x+7)^{28}(3x^2-4)+(6x)(30(x^3-4x+7)^{29})$$

Very quickly we can see we will have to continue doing chain rules and product rules every time we take a derivative. I also took the 3rd derivative, but the termas quickly expand and I do not see a pattern such that I can figure out what is supposed to happen at 100th derivative (especially because the chain rule and product rules really add terms). I calculated the answer online, and it is supposed to be zero, but this is not evident to me at all. Your help is much appreciated.

Answer 1

The highest power of $x$ in $g$ is $(x^3)^{30}=x^{90}$ which vanishes on the $91$st derivative. So, $$g^{(100)}(x)=0$$

Answer 2

If $g(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_0$, then differentiating it $(n+1)$-times gives zero. Now we have $$ g(x)=(x^3-4x+7)^{30}=x^{90} - 120x^{88} + 210x^{87} + \cdots + 22539340290692258087863249, $$ to be precise (we do not need the exact coefficients, of course).