What is the slope of the line normal to the tangent line of #f(x) = xsecx-cos(2x-pi/6) # at # x= (15pi)/8 #?
So first derive the equation, then plug in x into the derivative, then find the negative reciprocal of that slope to get the normal slope.
Use product rule to derive the first part:
Chain rule to derive second part:
Put these together so the derivative is
The slope of the normal (perpendicular) is the negative reciprocal of that slope, which will be