What is the derivative of #tan^-1(4x)#?

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Answer 1 · 2015-04-23T13:31:03

#4/(16x^2 +1)#

Derivative of #tan^-1 4x# can be written using the formula for the derivative of #tan^-1x# = #1/(1+x^2)#

#d/dx tan^-1 4x# = #1/(16x^2 +1) d/dx (4x)#

= #4/(16x^2 +1)#

Answer 2 · 2015-04-23T19:58:25

#y=arctan(4x)#

#tany=4x#

#sec^2y*(dy)/(dx)=4#

#(tan^2y+1)*(dy)/(dx)=4#

#(16x^2+1)*(dy)/(dx)=4#

#(dy)/(dx)=4/(16x^2+1)#