Question

How do you differentiate `arctan(x-(1+x^2)^(1/2))`?

Answer 1

`=[1/sqrt((x-(1+x^2)^(1/2))^2+1)][1-x(1+x^2)^(-1/2)]`

Explanation:

Before answering let us differentiate the inverse tan function.

`arctan(x)=y`
`tan(arctan(x))=tan(y)`
`d/dx[x]=d/dx[tan(y)]`
`1=sec^2(y)*(dy)/(dx)`
`1/(sec^2(arctanx))=(dy) /(dx)`

But `sqrt(tan^2(alpha)+1)=sec^2(alpha)`

Then in which `alpha=arctan(x)`:
`1/(sqrt(tan^2(arctan(x))+1))=(dy) /(dx)`
`1/(sqrt(x^2+1))=(dy)/(dx)`

Now differentiating:
`d/dx[arctan(x-(1+x^2)^(1/2))]`
`=[1/sqrt((x-(1+x^2)^(1/2))^2+1)]d/dx[x-(1+x^2)^(1/2)]`
`=[1/sqrt((x-(1+x^2)^(1/2))^2+1)][1-(1/2)(1+x^2)^(-1/2)d/dx[x^2+1]]`
`=[1/sqrt((x-(1+x^2)^(1/2))^2+1)][1-((1+x^2)^(-1/2))/2[2x]]`
`=[1/sqrt((x-(1+x^2)^(1/2))^2+1)][1-x(1+x^2)^(-1/2)]`