Answer
True
Work Step by Step
True.
We need the following fact.
$p(x)=x^n\Rightarrow p^{(n)}(x)=n!$ and $p^{(n+1)}(x)=p^{(n+2)}(x)=\ldots=0$
Given $f(x)=(x^6-x^4)^5$
Extending the polynomial,
$f(x)=x^{30}+\ldots-x^{20}$
Differentiating $30$ times and using the fact above,
$f^{(30)}(x)=30!+\ldots -0$
Differentiate once then we get
$f^{(31)}(x)=0$
