Answer
a) $f'(x) = cos(x)(1-6cos(x)sin(x))$
b) $f'(x) = sec(x)tan(x)(x^2-tan(x)) + (2x-sec^2(x))(1+sec(x))$
Work Step by Step
a) $f'(x) = cos(x) + 6 cos^2(x)(-sin(x)) = cos(x)(1-6cos(x)sin(x))$
by the chain rule.
b) $f'(x) = [1+sec(x)]'[x^2-tan(x)] + [1+sec(x)][x^2-tan(x)]' = sec(x)tan(x)(x^2-tan(x)) + (2x-sec^2(x))(1+sec(x))$
by the product rule.
