Question

find the derivative of tan^-1(2x)

Answer 1

\[\tan^{-1} (2x)\]

Answer 2

I"m told its suposed to be\[\frac{ 2 }{ 4x ^{2}+1 }\]

Answer 3

My problem is that i thought the formula for the inverse trig function was :

Answer 4

\[\frac{ 1 }{ x ^{2}+1 }\]

Answer 5

So by the chain rule wouldn't i take the derivative of tan and plug in 2x?

Answer 6

making it \[\frac{ 1 }{ 2x ^{2}+1 }\] ?

Answer 7

why is it 2 on the top and 4x^2 on the bottom?

Answer 8

y=arc tan 2x then by the chain rule you can find easily

Answer 9

\[\frac d{dx}\tan^{-1}u=\frac1{u^2+1}\cdot\frac{du}{dx}\]in your case, \(u=2x\)

Answer 10

yeah so it would be :

Answer 11

\[\frac{ 1 }{ 2x+1 } * 2 \]

Answer 12

if \(u=2x\) then \(u^2=?\)

Answer 13

which would give \[\frac{ 2 }{ 2x+1}\]

Answer 14

i meant the x to be squared on the bottom there

Answer 15

no, think it through...\[u=2x\implies u^2=(2x)^2=?\]

Answer 16

Oh. i was just putting the squared on the x and not also the two. thanks!

Answer 17

That makes much more sense now. Gave you a medal!

Answer 18

thanks, happy to help :)

Answer 19

if y=arc tan (2x) then tan y = 2x..