Ask your own question, for FREE!
Mathematics 79 Online
OpenStudy (dls):

Need help differentiating this at x=0

OpenStudy (dls):

\[\Huge y=\tan^{-1}(\frac{2^x}{1+2^{2x+1}})\]

OpenStudy (anonymous):

good lord

OpenStudy (dls):

Doing it directly is like..lol

OpenStudy (dls):

can we consider some substitution?I thought of 2^x=tan theta but wont work

OpenStudy (anonymous):

you don't really differentiate by substitution maybe a quick chain rule way? we can grind it out i suppose

OpenStudy (dls):

That would be hell lotta complicated

OpenStudy (anonymous):

interesting: a quick wolf check shows that \(\int \frac{2^x}{1+2^{2x+1}}\) involves arctan....

OpenStudy (anonymous):

we can try to grind it out, it can't be that hard

OpenStudy (dls):

hmm,I can do it directly,But I posted it so that someone could come up with a possibly easier method..

OpenStudy (anonymous):

i just checked via wolfram, and the derivative is not something snappy, so i don't think there is a quick gimmick to give the answer quickly http://www.wolframalpha.com/input/?i=arctan%282^x%2F%281%2B2^%282x%2B1%29%29%29

OpenStudy (dls):

@hartnn would you do it direct?

hartnn (hartnn):

let me try... tan y = 2^x /(1+2 . 2^[2x]) if we put x = log_2 u (same as u =2^x) tan y = u/ (1+2u^2) now find sec^2 y from here ...because we'll need it.... not sure whether this simplifies, just a thought

OpenStudy (dls):

I thought of muliplying and dividing 2 in num and denom

OpenStudy (dls):

it became 2tan theta somehow..if we could simplify denom?

hartnn (hartnn):

yeah, there is some formula , that i don't recollect now....tan^(-1) 2y = 2 tan^(-1)...something something..

hartnn (hartnn):

there must be some simplification happening.....

OpenStudy (dls):

I was just talking about tan2x's formula If we let 2^x=tantheta

OpenStudy (dls):

and multiply divide by 2 2.2^x ====2 tan theta and the denominator,if we could somehow make it 1+tan^2 x

hartnn (hartnn):

its, 2. 2^ (2x) = 2 tan^2 theta

OpenStudy (dls):

yeah,that's the only problem..

hartnn (hartnn):

you have the answer ?

OpenStudy (dls):

umm,yeah "none of these" :p

hartnn (hartnn):

dy/du = (2u^2-1) cos^2 y / (2u^2+1)^2 .... which is not difficult at x=0, u = 2^0 = 1, cos^2 {y} = cos^2 {tan^(-1) [1/3]} = 9/10 so, dy/du = (2-1) (9/10) (2+1)^2 = 1/10 phew

OpenStudy (dls):

what did you do?

hartnn (hartnn):

lastly, dy/dx = dy/du * du/dx ..... [u=2^x, du/dx = 2^x log 2] = 1/10 * log 2 = log 2 /10

hartnn (hartnn):

i did this : "tan y = 2^x /(1+2 . 2^[2x]) if we put x = log_2 u (same as u =2^x) tan y = u/ (1+2u^2)"

hartnn (hartnn):

then differentiated w.r.t u

OpenStudy (dls):

Options: 1 2 ln2 NOT

hartnn (hartnn):

but did u get what i did ?

OpenStudy (dls):

yeah,I did

hartnn (hartnn):

well, i checked my work again, could find any mistake, so i would go with NOT :P

OpenStudy (dls):

its NOT only :D

hartnn (hartnn):

hopefully ln 2/10 is also correct :P

OpenStudy (anonymous):

its easy.

OpenStudy (dls):

better method?

OpenStudy (anonymous):

|dw:1371557165590:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!