## kittycat01 2 years ago when u are having in this form then how to differentiate :) tan^-1(2^(x+1)/(1-4^x)) can i use substitution here ?

1. kittycat01

i don't think so substitution will work

2. hartnn

substitute 2^x = tan y

3. hartnn

lol, why not ?

4. hartnn

do you know tan 2 theta formula ?

5. kittycat01

coz i have never did this type of question lol

6. kittycat01

yep :)

7. kittycat01

2tanα/(1 – tan²α)

8. hartnn

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

9. kittycat01

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

10. kittycat01

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

11. hartnn

? what did you get after simplification ?

12. hartnn

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

13. kittycat01

wait :))

14. Abhishek619

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

15. kittycat01

@Abhishek619 :/

16. hartnn

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

17. Abhishek619

|dw:1372850429715:dw|

18. kittycat01

oh :O

19. kittycat01

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

20. hartnn

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

21. kittycat01

let me first @hartnn use ur method :D

22. kittycat01

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

23. hartnn

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

24. kittycat01

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

25. kittycat01

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

26. kittycat01

2tan^-1(2^x)

27. hartnn

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...

28. kittycat01

oh x is in power :O

29. kittycat01

then m getting the answer :)

30. hartnn

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

31. kittycat01

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

32. hartnn

welcome ^_^