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tan^-1[(√x+√a)/(1-√(xa))] differentiate it plz m gtting 1/2(√x) which is not correct answer is 1/[2(√x)(1+x)]

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same type of question with different substitution
hmm i think i did something wrong let me do it again
well u don't have to substitue :) @hartnn i think i forgot to differentiate tan^-1 (x)^1/2 i missed the tan^-1 part

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Other answers:

say, sqrt x= tan y sqrt a = tan alpha then , [(√x+√a)/(1-√(xa))] =... ? now did u get it ?
u have to use tan^-1 x + tan^-1 y formula well not now
trying again
:'( not getting :'(
how to differnatiate tan^-1(x)^1/2
use chain rule, know about it ?
no :'( that's the problem simple chain rule i know
like this one sin(cos3x) but never did for inverse trign.........
is the differenatition of tan^-1(x)^1/2 is
is =X =1/(1+(x^1/2)^2 . 1/2*(x)^-1/2 ????????
ok, derivative of tan^{-1} x is 1/(1+x^2), right ?? for derivative of tan^{-1} sqrt(x) you do 1/(1+ [\sqrt x]^2) times deribative of sqrt x
i think i have used chain rule -_- is that a chain rule @hartnn
you did it,,,kitty
didn't you get the answer then ? yes, thats was chain rule only...
lol i know everything but still do silly things -_- @souvik
it happens some time...
@hartnn - chain rule :O so that's a chain rule
because what you wrote matches with your final answer...
not some times everytime it happens
keep your head MSD
bcoz of calculation mistake i lost 3 marks in maths prob :'( i did everything correct but in final answer i made a error in multiplication
@hartnn -tahnx :)
hey, welcome anytime ^_^
Indian cricket team captain...
oh :OOO ms dhoni :OOO i don't like him lol
reminds me of my frenemy :P

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