How do you find an equation of the tangent line to the curve at the given point #y=tan^2(3x)# and #x=pi/4#?

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Answer 1 · 2017-10-14T12:54:53

The equation of the tangent at #(pi/4,1)# is #y= -12x +(3pi+1)#

#y=tan^2(3x)# at #x=pi/4 ; y = tan^2(3*pi/4) =(-1)^2=1 :.#

at #(pi/4,1)# the tangent is drawn.

#:. y^'=2tan(3x)*sec^2(3x).3= 6tan(3x)*sec^2(3x) #

at #x=pi/4 , y^'= 6tan((3pi)/4)*sec^2((3pi)/4) # or

#y^'= 6* (-1)*(-sqrt2)^2= -12 :. # At #x=pi/4#, the slope of

the tangent is #m=-12 :.# the equation of the tangent at

#(pi/4,1)# is #y-y_1=m(x-x_1) or y-1= -12(x-pi/4)# or

# y= -12(x-pi/4) +1 or y= -12x +(3pi+1)# [Ans]