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

NetherCraft 0

1 Answer

  • y = 2 ln(secx)

    Using the chain rule,

    d(secx)/dx = secx tanx

    d(ln(x))/dx = 1/x,


    y’ = 2 secx tanx / secx;


    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.

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