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.