Question

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

Answer 1

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

Explanation:

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

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

Explanation:

Let

`y=tan^-1x`

Then,

`tany=x`

Implicit differentiation yields

`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)`