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calculus

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Find an equation of the tangent line to the curve at the given point.
y = 7 sec x

P = (π/3, 14)

  • calculus - ,

    y(x)=7sec(x)
    y'(x)=7sec(x)tan(x)

    at x=π/3,
    slope of tangent, m =
    y'(x)= 7*sec(π/3)tan(π/3)
    =7*2*sqrt(3)
    =14sqrt(3)

    We now look for a line passing through P1(x0,y0)=(π/3,14) with a slope of m=14sqrt(3).

    The equation to use is:
    (y-y0)=m(x-x0)
    substitute x0=π/3, y0=14, m=14sqrt(3) and simplify to get equation of tangent.

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