1 Answer

y = 2 ln(secx)
Using the chain rule,
d(secx)/dx = secx tanx
d(ln(x))/dx = 1/x,
Therefore,
y’ = 2 secx tanx / secx;
or
y’ = 2 tanx.
tan(π/4) = 1, so
y'(π/4) = 2 * 1 = 2.
This is the slope of the tangent line at x = π/4. The normal is perpendicular to the tangent, so the slope would be the negative reciprocal. The slope of the normal at x = π/4 is therefore 1/2.