1. Find the slope of the function y(x) = 2 cos(x) sin(2x) at x=π/2
y'(x)=-2sin(x)sin(2x)+4cos(x)cos(2x), which at x=π/2 is 0
since sin(π)=0 and cos(π)=0
2. Find the equation of the line that is tangent to y=x2sin(2x) at x=-π
y'(x)=2xsin(2x)+2x2cos(2x)
y(-π)=0, y'(-π)=-2πsin(-4π)+2π2cos(-4π)=2π2
The line has the equation y=2π2x+b, which, for a line through the point (-π,0), of that slope
gives b=2π3, so have y=2π2x+2π3
3.Determine the second derivative for y=x2cosx
y'(x)=2xcosx-x2sinx
y''(x)=2cosx-2xsinx-2xsinx-x2cosx=(2-x2)cosx-4xsinx