What is the derivative of inverse tangent of 2x?

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Answer 1 · 2015-04-14T14:38:10

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

Solution
Let

#y=tan^(-1)2x#

#tany=2x#

Differentiating both side with respect to 'x'

#d/dx(tany)=d/dx(2x)#

#=>sec^2y(dy/(dx))=2#

#=>dy/(dx)=2/(sec^2y)#

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

Now, as
#tany=2x#

#tan^2y=(2x)^2#

#tan^2y=4x^2#

So,

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

Answer 2 · 2015-04-14T14:44:41

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

Solution

Let

#y=tan^(-1)2x#

Differentiating both side with respect to 'x'

#dy/dx=d/dx(tan^(-1)2x)#

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

#dy/dx=1/(1+(2x)^2).2d/dx(x)#

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

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

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