SOLUTION: If y = tan^-1(x/a) So y'=a/(x^2+a^2) [by differentiating] Y'=a/(x-ia)(x+ia) why does it become like this and where does the i comes from? Sir, this question is from calcu

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: If y = tan^-1(x/a) So y'=a/(x^2+a^2) [by differentiating] Y'=a/(x-ia)(x+ia) why does it become like this and where does the i comes from? Sir, this question is from calcu      Log On


   

Question 891484: If y = tan^-1(x/a)
So y'=a/(x^2+a^2) [by differentiating]
Y'=a/(x-ia)(x+ia) why does it become like this and where does the i comes from?
Sir, this question is from calculus (examples of finding nth derivative so please help me.)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
When you look at the difference of two squares,
x%5E2-a%5E2=%28x%2Ba%29%28x-a%29
since,
%28x%2Ba%29%28x-a%29=x2-ax%2Bax-a%5E2 so the x terms cancel out.
However when you look at the sum of two squares,
x%5E2%2Ba%5E2 you have the same situation with trying to cancel out the x terms, however if you look at %28x%2Ba%29%28x%2Ba%29 you get,
%28x%2Ba%29%28x%2Ba%29=x%5E2%2B2ax%2Ba%5E2} so that won't work.
You need the opposite signs to cancel out the x terms and you need the a%5E2 term to have a negative coefficient.
Since i%5E2=-1 then you can use i as the coefficient.
%28x-ai%29%28x%2Bai%29=x%5E2%2Baix-aix%2B%28ai%29%5E2
%28x-ai%29%28x%2Bai%29=x%5E2-a%5E2