#f(x) = xsinx-cos^2x#

Apply product rule, chain rule and standard derivatives.

#f'(x) = xcosx+sinx- 2cosx*(-sinx)#

#= xcosx+sinx+2sinxcosx#

#= xcosx+sinx+sin(2x)#

The slope of #f(x)# at #x=pi/8# is #f'(pi/8)#

#f'(pi/8) = pi/8*cos(pi/8) +sin(pi/8) + sin(pi/4)#

#approx 1.542597# (Calculator)

The slope of a tangent (#m_1#) #xx# slope of a normal (#m_2#) at any point on a curve is given by: #m_1*m_2 =-1#

#:.# Slope of the normal to #f(x)# at #x=pi/8 approx -1/1.542597#

#approx -0.688422#

The equation of a straight line of slope #m# passing through the point #(x_1,y_1)# is given by:

#(y-y_1) = m(x-x_1)#

In this example, our normal will have the equation:

#y-f(pi/8) approx -0.688422 (x-pi/8)#

#y- (-0.703274) approx -0.688422 (x- 0.392699)# (Calculator)

#y+0703274 approx -0.688422x+0.270343#

#y approx -0.688422x-0.432931#

#y =-0.69x-0.43# [To 2D]

We can see #f(x)# and the normal at #x=pi/8# on the graphic below.