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Find the equation of the tangent line to the curve y=6secx−12cosx at the point (π/3,6). The equation of this tangent line can be written in the form y=mx+b where
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i get how to get m but not b
b is the point where line crosses the y-axis (where x = 0)
i know that but dont know how to find it
what is the slope?
use the form y-y1 = m(x-x1) where m = slope , x1 =pi/3 and y1 = 6 and rearrangge to y = mx + b
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y-6=(6 (cos(2 x)+2) tan(x) sec(x))(x-(pi/3)
confused on where i go from here
y = ( 6(cos(2 x)+2) tan(x) sec(x)) * x - [6 (cos(2 x)+2) tan(x) sec(x))*(pi/3)] - 6
oopps + 6 at the end
we can simplify the 'b' a little - 6 [ (cos(2 x)+2) tan(x) sec(x))(pi/3) - 1]
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