Find the $2022$th derivative $f^{(2022)}(0)$ of the function $f(x) = x^{2019}\cos(x)$. Find $f^{(2022)}(0)$  of the function
$$f(x)=x^{2019} \cos x.$$
By Taylor series at the point $x =0$, my answer was $0$.
But someone who gave this question to me said the answer is like the below picture.
Who is correct?
Did I something wrong?


You are correct; the coefficient of $x^{2022}$ in the Taylor series of $x^{2019}\cos x$ at $x=0$ is $0$, which implies that the 2022th derivative at $0$ is $0$.
Your friend made the following error. Yes, all the terms but the $k=2019$th one dissapear. But this means s/he should have $\cos^{(2022-2019)}(0) = \cos'''(0) = -\sin(0) = 0.$
