when u are having in this form then how to differentiate :) tan^-1(2^(x+1)/(1-4^x)) can i use substitution here ?

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when u are having in this form then how to differentiate :) tan^-1(2^(x+1)/(1-4^x)) can i use substitution here ?

Mathematics
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i don't think so substitution will work
substitute 2^x = tan y
lol, why not ?

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Other answers:

do you know tan 2 theta formula ?
coz i have never did this type of question lol
yep :)
2tanα/(1 – tan²α)
2^(x+1) = 2. 2^x , 4^x = (2^x)^2 plug in 2^x = tan y and see the magic! (simplification :P)
will it really works :PPPP lol i hate chain rule -_-
don't wanna use chain rule for inverse trig ...........
? what did you get after simplification ?
i don't think chain rule is required here...
wait :))
use tan(x+y)=(tanx+tany)/(1-tanxtany)
same thing....tan 2 thetha comes from tan (x+y) formula only
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oh :O
@Abhishek619 -oh :OO yeah we can do it in that way also :OOO tahnx :)
with substitution, you would have got the same thing...
let me first @hartnn use ur method :D
@Abhishek619 @hartnn -answer is [2^(x+1)loge2]/(1+4^x) :'(
i am never interested in final answer, just the method..... if your algebra part is correct, you WILL get correct final answer....
but the answer is in loge2 and having (1+4^x) also
well did he wrote 2tan^-12^x?????????
2tan^-1(2^x)
ok, got till 2 tan inverse 2^x, right ? if you diff. that u get 2 * (1/1+(2^x)^2) 2^x ln 2 (ln = log_e) .... (2^x ln 2 comes from chain rule) yes...
oh x is in power :O
then m getting the answer :)
yeah... (2^x)^2 = 4^x and 2.2^x = 2^(x+1) good!
@hartnn -thanx :) now i can do the rest ^_^
welcome ^_^

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