Question

First derivative of (1+x^2) Arctan x-x

Answer 1

What?

Answer 2

\[\Large\bf\sf \frac{d}{dx}(1+x^2)\arctan x-x\]

Answer 3

What's being operated upon?

Answer 4

inverse trigo

Answer 5

the answer was 2x arctan x.. and still, no idea deriving the right answer

Answer 6

We need the right question to give the right answer.

Answer 7

\(\Large\bf\sf \frac{d}{dx}(1+x^2)\arctan x-x\)

does give 2x arctan x

Answer 8

did u try product rule for first term

Answer 9

what do you mean?

Answer 10

ah! yes

Answer 11

heard of product rule ?
\(\huge (uv)' = u'v+uv'\)

here
u= 1+x^2
v = arctan x

there you go :)

Answer 12

i got 2x arctan x-x.. and.. it wasnt x-x,

Answer 13

how come you get the 2nd term as -x ??
shouldn't it be -1 ?

Answer 14

You guys should really be more clear about the derivatives.

Answer 15

i mean +1

Answer 16

what i did first was (1+x^2) (x-x dy/dx / 1-2x^2) + (2x) (arc tan x-x)