the slope of the line normal to the graph of y=2Ln(secx) at x=pie/4 is?

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.