Derivative of \(\displaystyle{f{{\left({x}\right)}}}={\arctan{{\left(\sqrt{{{\frac{{{1}+{x}}}{{{1}-{x}}}}}}\right)}}}\) I am trying to find

Nancy Richmond

Nancy Richmond

Answered question

2022-03-28

Derivative of f(x)=arctan(1+x1x)
I am trying to find the derivative of f(x)=arctan(1+x1x) by only using the formula arctan(u(x))=u(x)u(x)2+1. I don't honestly understand this formula, here are however my calculations. Is this the right way to find the derivative? And how do I proceed?
121+x1x(1x)(1+x)x22x+1x+11x+1
I also appreciate if you can explain what that "formula" exactly is.

Answer & Explanation

Drake Huang

Drake Huang

Beginner2022-03-29Added 15 answers

Well done! Your answer is correct. To make it a bit more presentable, let us simplify this further to get:
f(x)=1x1+x2×2xx22x+12(1x)
=(1x)3221+x(x22x+1)
=121x2
EDIT:
Well, you have your derivative calculated wrong, it should be
(1x)(1+x)(1)x22x+1=2x22x+1
So, removing the minus sign, we have:
f(x)=121x2

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