Question

Derive arcsin e^(arctan x).

Answer 1

\[\huge \sin^{-1}(e^{\tan^{-1} x})\]
is that the question?

Answer 2

Yes it is :)

Answer 3

haha nice...i'll try it :DDD

Answer 4

I have a very messy equation at the moment. very ugly :L

Answer 5

first let \[\huge y = \sin^{-1} (e^{\tan^{-1} x})\]
take the sin of both sidees
\[\huge \sin y = e^{\tan^{-1} x}\]
implicitly differentiat
\[\huge (\cos y) y' = (e^{\tan^{-1} x})\frac{d}{dx} (\tan^{-1}x)\]
does that help?

Answer 6

I don't get its relevance though...

Answer 7

what do you mean relevance?

Answer 8

Urg

Answer 9

a possible way is also this:
\[\huge \sin y = e^{\tan^{-1} x}\]
ln both sides
\[\huge \ln (\sin y) = \tan^{-1} x\]
tangent both sides
\[\huge \tan (\ln |\sin y| ) = x\]
implicitly differentiate :D

Answer 10

I reached \[e^{\tan^{-1}x}\div((1+x^{2})(\cos y))\]

Answer 11

yes sounds right

Answer 12

Is there a way to simplify?

Answer 13

now i guess you need to express cos y in terms of x

Answer 14

hmm in times of doubt//time to consult wolfram =_=

Answer 15

I'm just grinding out the answer .-.

Answer 16

I'm taking AP calc BC, its hard as heck.

Answer 17

OHHH i got it

Answer 18

i thought it was absurd when i thought it at first but heh

Answer 19

\[\huge \sin y = e^{\tan^{-1} x} \]

Answer 20

so that means the other side must be