What is the equation of the tangent line of #f(x)=6tanx # at #x=pi/4#?

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Answer 1 · 2017-01-11T05:29:52

#y=12x-3pi-6=12x=3.425#, nearly. The point of cotact is #(pi/4, 6)#.

graph{(y-6 tan x)(y-12x+3.42)=0 [-1.57, 1.57, -10, 100]} Slope #f' = 6 sec^2x = 12, at x = pi/4#.

The equation of the tangent is

#y-6 tan(pi/4)=12(x-pi/4)# that simplifies to

#y=12x-3pi-6=12x=3.425#, nearly

See the Socratic tangent-inclusive graph for one period [#-pi/2, pi/2]#.

The period of the graph is #pi#.

The periodic graph is also given

graph{(y-6 tan x)(y-12x+3.4)=0 [-50, 50, -25, 25]}