Question

What is the slope of the line normal to the tangent line of `f(x) = xsecx-cos(2x-pi/6)` at `x= (15pi)/8`?

Answer 1

0.2865...

Explanation:

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:
`d/dx[xsec(x)]= sec(x)+ xtan(x)sec(x)`
Chain rule to derive second part:
`d/dx[-cos(2x-pi/6)]=2sin(2x-pi/6)`
Put these together so the derivative is
`f'(x)= sec(x)+ xtan(x)sec(x)+2sin(2x-pi/6)`
Plug in `x=(15pi)/8` to get the slope of the parallel line as `-3.49041`.
The slope of the normal (perpendicular) is the negative reciprocal of that slope, which will be `0.2865`