How do you take the derivative of #tan^-1(x)#?

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Answer 1 · 2018-07-30T11:05:32

# d/(dx)(tan^-1x)=1/(1+x^2) ,x inRR#

We know that ,

#(f^-1)'(x)=1/(f'(f^-1(x))) ,f'(x)!=0#

OR

#(dy)/(dx)=1/((dx)/(dy))#

Let ,

#tan^-1x=arc tanx=y# , where , #y in(-pi/2,pi/2)#

So , #x=tany# , where , #x in RR#

Diff. #x=tany# w.r.t. #y#

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

Subst. #tany=x#

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

So,

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

#:. d/(dx)(tan^-1x)=1/(1+x^2) ,x inRR#

Answer 2 · 2018-07-30T11:19:41

The answer is #=1/(1+x^2)#

Let

#y=tan^-1x#

Then,

#tany=x#

#sec^2ydy/dx=1#

#dy/dx=1/(sec^2y)#

But

#1+tan^2y=sec^2y#

#sec^2y=1+x^2#

Finally,

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