Posted by sh on Wednesday, March 24, 2010 at 12:24am.
Find an equation of the tangent line to the given curve at the given point
y=(1/cosx)2cosx at((pi/3),1)
y'=0+2sinx
slope = √3
y=mx+b
1=(√3)(pi/3)+b
b=(3√3pi)/3
3√3x3y+3√3pi=0
What did I do wrong? Thanks in advance.

Calculus  sh, Wednesday, March 24, 2010 at 1:20am
Got it.
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