Determine the derivative for y=tanxcosx

Question

a) y'=-sinxtanx + cosxsec^2x

b) y'=-sinxtanx + cosxtanx

c) y'=cosxsec^2x

d) y'=sinx - cosxsec^2x

Answer 1

Most of the above use the product rule,
(d/dx)u.v=u(dv/dx)+v(du/dx)
For example:
y=-sin(x)tan(x)
y'=-sin(x)d(tan(x))/dx + tan(x)d(-sin(x))/dx
=-sin(x)sec²(x)-tan(x)cos(x)
=-sin(x)/cos²(x)-sin(x)
=-sin(x)(1+sec²(x))

Answer 2

A is your answer because what you have there is the product rule[F'(x)*G(x)+G'(x)*F(x)].