Question

What is the slope of the line normal to the tangent line of `f(x) = sec^2x-xcos(x-pi/4)` at `x= (7pi)/4`?

Answer 1

The reqd. slope = `4/((16+7pi)`.

Explanation:

`f(x)=sec^2x-xcos(x-pi/4)`
`:.f'(x)=2*secx*secx*tanx-{x*(-sin(x-pi/4))+cos(x-pi/4)} = 2sec^2x*tanx+x*sin(x-pi/4)-cos(x-pi/4)`

But `f'(x)` is the slope tangent (tgt.) line at `(x, f(x))`
Hence, slope of the tangent line at `x=(7pi)/4` is
`f'((7pi)/4)=2sec^2((7pi)/4)tan((7pi)/4)+((7pi)/4)sin((7pi)/4-pi/4)-cos((7pi)/4-pi/4)=2*2*(-1)+((7pi)/4)*(-1)-0=-4-(7pi)/4=-(16+7pi)/4`
As normal is perpendicular to tgt. line, the reqd. slope = `4/((16+7pi)`.