How do you find the derivative of #y=tan(3x)#?

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Answer 1 · 2018-04-14T16:05:32

#(dy)/(dx)=3sec^2 3x#

we use teh chain rule

#(dy)/(dx)=color(red)((dy)/(du))(du)/(dx)#

#y=tan3x#

#u=3x=>(du)/(dx)=3#

#:color(red)(y=tanu=>(dy)/(du)=sec^2u)#

#(dy)/(dx)=color(red)(sec^2u)xx3#

#(dy)/(dx)=3sec^2u=3sec^2 3x#