What is the derivative of #tan(2x)#?

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Answer 1 · 2016-10-26T15:50:35

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

#y=tan(2x)#

# u=2x=>(du)/dx=2#

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

#(dy)/(dx)=(dy)/(du)xx(du)/(dx)#

#:.(dy)/(dx)=(sec^2u)xx2#

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