What is the equation of the tangent line of #f(x) = sec^3x^2# at #x=pi/8#?

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Answer 1 · 2017-07-20T19:43:07

#y = (3pi)/4sec^3(pi^2/64)tan(pi^2/64)x-(3pi^2)/32sec^3(pi^2/64)tan(pi^2/64)+ sec^3(pi^2/64)#

At #x = pi/8#, we have #y = f(pi/8) = sec^3(pi^2/64)#

And the slope is given by

#f'(x) = 3sec^2(x^2) sec(x^2) tan(x^2) * 2x#

# = 6xsec^3(x^2)tan(x^2)#

At #x = pi/8#, the slope of the tangent line is

#m = f'(pi/8) = 6pi/8sec^3(pi^2/64)tan(pi^2/64)#

# = (3pi)/4sec^3(pi^2/64)tan(pi^2/64)#

the equation of the line through #(pi/8, sec^3(pi^2/64))# with slope #m = (3pi)/4sec^3(pi^2/64)tan(pi^2/64)# is

#y = (3pi)/4sec^3(pi^2/64)tan(pi^2/64)x-(3pi^2)/32sec^3(pi^2/64)tan(pi^2/64)+ sec^3(pi^2/64)#

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