We will first find the derivative of #f(x)#.
#[1]" "f'(x)=d/dx(x^2cos2x)#
By product rule, this will become:
#[2]" "f'(x)=d/dx(x^2)*cos2x+d/dx(cos2x)*x^2#
By power rule, #d/dx(x^2)# becomes #2x^1#.
#[3]" "f'(x)=2x*cos2x+d/dx(cos2x)*x^2#
The derivative of #cosx# is #-sinx#. But since we are getting the derivative of #cos2x# instead, we need to use chain rule.
#[4]" "f'(x)=2x*cos2x+(-sin2x)*d/dx(2x)*x^2#
The derivative of #2x# is simply #2#.
#[5]" "f'(x)=2x*cos2x+(-sin2x)*2*x^2#
#[6]" "f'(x)=2xcos2x-2x^2sin2x#
Now that we have the derivative of #f(x)#, we can solve for the slope of the tangent line at #x=pi/4#.
#[1]" "m=2(pi/4)cos(2*pi/4)-2(pi/4)^2sin(2*pi/4)#
#[2]" "m=(pi/2)cos(pi/2)-(pi^2/8)sin(pi/2)#
#[3]" "m=(pi/2)(0)-(pi^2/8)(1)#
#[4]" "m=-pi^2/8#
We now look for the point of #f(x)# at #x=pi/4#.
#[1]" "f(pi/4)=(pi/4)^2cos(pi/2)#
#[2]" "f(pi/4)=(pi^2/16)(0)#
#[3]" "f(pi/4)=0#
#[4]" "(x_0,y_0)=(pi/4,0)#
Finally, we can solve for the equation of the tangent line.
#[1]" "y-y_0=m(x-x_0)#
#[2]" "y-0=(-pi^2/8)(x-pi/4)#
#[3]" "color(red)(y=(-pi^2/8)x+pi^3/32)#
