Question

Differentiate please.. !

Answer 1

\[\tan^{-1} (\frac{\sqrt{1+x^2}-1}{x}) w.r.t \tan^{-1}x\]

Answer 2

x=tan theta

Answer 3

it should be 1/2 if you do it correctly

Answer 4

its dam simple if you substitute x=tan theta

Answer 5

@DLS I applied your suggestion, and get 1/ sqrt (1 -x^2), not 1/2 .

Answer 6

you will get 1/2 arctanx on simpliying this

Answer 7

and u have to differentiate WRT arctanx so it will cancel out

Answer 8

The function becomes arctan( sin y - cot y)

Answer 9

ok, please, watch me

Answer 10

x = tan theta, --> theta = arctan x ,

Answer 11

sqrt (1 +tan^2 ) -1 / x = sqrt (sec^2 ) -1/tan x
= sec -1 /tanx

Answer 12

and then?

Answer 13

take quotient differentiate?

Answer 14

NONONONOONONOO

Answer 15

yes, guide me please

Answer 16

\[\Huge y=\frac{\sec \theta -1}{\frac{\sin \theta}{\cos \theta}}\]
\[\Huge y=\frac{1-\cos \theta}{\sin \theta}\]
\[\Huge y=\frac{1-(1-{2 \sin^2 \frac{\theta}{2}})}{2\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\]
\[\Huge y=\tan^{-1}(\tan \frac{\theta}{2})\]
\[\Huge y=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x \]
for all x(-pi/2,pi/2)

Answer 17

thank you very much.

Answer 18

yw :)