Question

y=xSquare secx

Answer 1

differentiate again?

Answer 2

same question

Answer 3

yes

Answer 4

product rule...udv + vdu

x^2(secxtanx) + 2xsecx

Answer 5

\[y=x^2\sec(x)\]?

Answer 6

use
\[(fg)"=f'g+g'f\] with
\[f(x)=x^2, f'(x)=2x, g(x)=\sec(x), g'(x)=\sec(x)\tan(x)\] and plug right in