Question

find the derivative of y = arctan(x/9) − 1/(2(x2 + 81) )

Answer 1

d/dx of arctan(x/9) = \[\frac{\frac{1}{9}}{1+(\frac{x}{9})^2}\]

Answer 2

\[\frac{x}{(x^2+81)^2}\] for the other one...

Answer 3

looks good

Answer 4

that can't be the answer tho..

Answer 5

why not?

Answer 6

Idk maybe it cna, I was self checking with wolfman and it's dioofferent....

http://www.wolframalpha.com/input/?i=find+the+derivative+of+y+%3D+arctan%28x%2F9%29++%E2%88%92++1%2F%282%28x2+%2B+81%29++%29

Answer 7

it's just some algebra; a different way of writing the same thing

Answer 8

try simplifying the first expression

Answer 9

it could be 1+(x^2/81?)

Answer 10

1/9/that :P

Answer 11

\[\frac{\frac{1}{9}}{1+(\frac{x}{9})^2}\cdot\frac99=\frac1{9+9x^2}\]so we have\[\frac1{9+9x^2}+\frac x{(x^2+81)^2}\]then we gotta rationalize that

Answer 12

Happy 100th anniversary Turing!

Answer 13

thanks, lol

Answer 14

turning is 100? no wonder :P.

Answer 15

ooh I should have multiplied by 9^2=81 I think\[\frac{81}{81+{x^2}}+\frac x{(x^2+81)^2}={(x^2+81)^2+x(81+x^2)\over(81+x^2)^2}\]

Answer 16

with me so far?

Answer 17

mhmhm

Answer 18

we want the denominator on the first to look like that of the other, and we can do that by multiplying by \(\frac{9^2}{9^2}=\frac{81}{81}\)\[{\frac19\over1+(\frac x9)^2}\cdot\frac{81}{81}={9\over81+x^2}\]oh, I made a mistake earlier...
so that is the first term

Answer 19

extra hairy....

Answer 20

I think I got it from the wolfman explanation... I don't want you to waste your time :). Thanks!

Answer 21

so now we have\[{9\over81+x^2}+{x\over(x^2+81)^2}\]

Answer 22

welcome!

Answer 23

onto the next questions hehehe :).