At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

i don't think so substitution will work

substitute 2^x = tan y

lol, why not ?

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

|dw:1372850429715:dw|

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...

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

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

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

2tan^-1(2^x)

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!

welcome ^_^