# How do you differentiate f(x)=sec^4(x^3-x^2 )  using the chain rule?

$f ' \left(x\right) = 4 \left(3 {x}^{2} - 2 x\right) {\sec}^{4} \left({x}^{3} - {x}^{2}\right) \tan \left({x}^{3} - {x}^{2}\right)$
$f \left(x\right) = {\sec}^{4} \left({x}^{3} - {x}^{2}\right) = {\left(\sec \left({x}^{3} - {x}^{2}\right)\right)}^{4}$
f'(x)=4(sec(x^3-x^2))^3*sec(x^3-x^2))tan(x^3-x^2)*(3x^2-2x)
$f ' \left(x\right) = 4 \left(3 {x}^{2} - 2 x\right) {\sec}^{4} \left({x}^{3} - {x}^{2}\right) \tan \left({x}^{3} - {x}^{2}\right)$