Question

Trig derivative

Answer 1

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

Answer 2

Should have said inverse trig

Answer 3

Derivative of arctan on it's own is:
\[\frac{ 1 }{ x^{2} + 1 }\]So that is your outer layer for the chain rule, now just multiply that by the derivative of the angle.

Answer 4

yeah I did all that but my final answer is wrong, nots ure where i messed up haha

Answer 5

\[\frac{ 1-(1/2)(1+x^{2})^{-1/2}(2x) }{ x^{2}+1 }\]
\[\frac{ 1-\frac{ x }{ (x^{2}+1)^{1/2} } }{ x^{2}+1 }\]
\[\frac{ \frac{\sqrt{x^{2}+1}- x}{\sqrt{x^{2}+1}} }{ x^{2}+1 } \implies \frac{ \sqrt{x^{2}+1}- x }{ (x^{2}+1)^{3/2} }\]

Dunno, how much do we have to simplify this? xD

Answer 6

mhm yeah I think I messed up on the second last step, makes sense.

Answer 7

Thank you again ^.^

Answer 8

Its not calculus thats hard, its algebra xD

Answer 9

Right?

Answer 10

I always mess up with my algebra O_O

Answer 11

Everyone does xD Guess just a lot of practice with this stuff. Until you get used to this specific kind of algebra, its easy to screw up. I know I do.

Answer 12

No no you're good at everything :D

Answer 13

i mess up more often than Id like, lol.

Answer 14

http://www.youtube.com/watch?v=LEsIfAWRtnA every time you answer. Replace bill murry with your name haha

Answer 15

xD random, lol.

Answer 16

Ik lool