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July 28, 2014

July 28, 2014

Posted by **Anonymous** on Sunday, February 27, 2011 at 8:25pm.

y = 7 sec x

P = (π/3, 14)

- calculus -
**MathMate**, Sunday, February 27, 2011 at 9:15pmy(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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