if g(x)=x^3coz2x find g'(0)

Question

πThere is two questions. The first is find g'(pi) and the second is find g'(0).

Thanks for your help! I've looked through all of my notes and just can't seem to figure out how to do them.

Answer 1

g(x) = x³cos2x

 

Use the product rule, and apply the chain rule when you differentiate cos2x.

 

g'(x) = (3x²)cos2x + x³(-2 sin2x) = 3x²cos2x - 2x³sin2x

 

By inspection, g'(0) = 0

 

g'(π) = 3π²cos(2π) - 2π³sin(2π) = 3π²(1) - 2π³(0) = 3π²