Question

obtain dy/dx:
y=sec^3(x-x^2)^4

Answer 1

\[\large y = \sec^3\left(x-x^2\right)^4 = \large y = \left(\sec\left(x-x^2\right)^4\right)^3\]

Answer 2

use chain rule

Answer 3

\[ \large y' = 3\left(\sec\left(x-x^2\right)^4\right)^{3-1}\left(\sec\left(x-x^2\right)^4\right)'\]

Answer 4

keep unfolding using chain rule..