kittycat01
when u are having in this form then how to differentiate :)
tan^1(2^(x+1)/(14^x))
can i use substitution here ?



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kittycat01
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i don't think so substitution will work

hartnn
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substitute 2^x = tan y

hartnn
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lol, why not ?

hartnn
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do you know tan 2 theta formula ?

kittycat01
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coz i have never did this type of question lol

kittycat01
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yep :)

kittycat01
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2tanα/(1 – tan²α)

hartnn
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2^(x+1) = 2. 2^x , 4^x = (2^x)^2
plug in 2^x = tan y and see the magic! (simplification :P)

kittycat01
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will it really works :PPPP
lol i hate chain rule _

kittycat01
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don't wanna use chain rule for inverse trig ...........

hartnn
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?
what did you get after simplification ?

hartnn
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i don't think chain rule is required here...

kittycat01
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wait :))

Abhishek619
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use tan(x+y)=(tanx+tany)/(1tanxtany)

kittycat01
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@Abhishek619 :/

hartnn
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same thing....tan 2 thetha comes from tan (x+y) formula only

Abhishek619
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dw:1372850429715:dw

kittycat01
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oh :O

kittycat01
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@Abhishek619 oh :OO
yeah we can do it in that way also :OOO
tahnx :)

hartnn
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with substitution, you would have got the same thing...

kittycat01
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let me first @hartnn use ur method :D

kittycat01
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@Abhishek619 @hartnn answer is [2^(x+1)loge2]/(1+4^x) :'(

hartnn
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i am never interested in final answer, just the method.....
if your algebra part is correct, you WILL get correct final answer....

kittycat01
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but the answer is in loge2 and having (1+4^x) also

kittycat01
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well did he wrote 2tan^12^x?????????

kittycat01
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2tan^1(2^x)

hartnn
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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...

kittycat01
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oh x is in power :O

kittycat01
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then m getting the answer :)

hartnn
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yeah...
(2^x)^2 = 4^x
and 2.2^x = 2^(x+1)
good!

kittycat01
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@hartnn thanx :)
now i can do the rest ^_^

hartnn
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welcome ^_^