If g(x)=sec(x), find g'''(5pi/3)?

Question

Thank you so much!

Answer 1

g(x) = sec(x)

g'(x) = sec(x)tan(x)

--> [ sec(x) [ sec^2(x)] ] + [ tan(x) [ sec(x)tan(x) ] ]
-->
--> [ sec^3(x) ] + [ sec(x)tan^2(x) ]

g''(x) = sec^3(x) + sec(x)tan^2(x)

--> [ 3 sec^2(x) [ sec(x)tan(x) ] ] + [ sec(x) [ 2 tan(x) [ sec^2(x)] ]
----> + tan^2(x) [ sec(x)tan(x) ] ]
-->
--> [ 3 sec^3(x)tan(x) ] + [ 2sec^3(x)tan(x) + sec(x)tan^3(x) ]
-->
--> [ 5 sec^3(x)tan(x) + sec(x)tan^3(x)

g'''(x) = 5sec^3(x)tan(x) + sec(x)tan^3(x)

g'''(5π/3) = 5(8)(-1.732) + (2)(-5.196)

g'''(5π/3) = 109.608 (rounded)