Question

Please help, explanation would be greatly appreciated, picture included

Answer 1

First You need to find the derivative of y = arctan (x/3)
this gives the slope of the tangent

Answer 2

can you do that?

Answer 3

y'=x/(x^2+9) is what i got

Answer 4

i dont think thats correct
let me check..

Answer 5

No its 3 / (x^2 + 9)

Answer 6

so at the origin (0,0) the slope is 3/9 = 1/3

Answer 7

do we use quotient rule for x/1?

Answer 8

No i just used the standard derivative for arctan (x / a)

Answer 9

y' = a / (a^2 + x^2)

Answer 10

i haven't learned that one yet i just know arctanu= u'/ (1+u^2)

Answer 11

- its basically the same thing

Answer 12

i understand where that comes from
because u is over 1

Answer 13

yes

Answer 14

ok so what do we do once we have the correct derivative?

Answer 15

y'=3/(x^2+9)

Answer 16

well the graph passes through (0,0) and the slope = 1/3

so using
y - y1 = m (x - x1)


where m = 1/3 and x1 = 0 and y1 = 0 we have

y = (1/3)x

or 3y = x

or x - 3y = 0

Answer 17

can you please just give me the quick way you found the slope please

Answer 18

slope = 3 / (x^2 + 3^2)
at the origin x = 0 so
m = slope = 3 / (0 +9) = 1/3

Answer 19

i use the standard derivative of arctan (x/a) = a / (a^2 + x^2)

Answer 20

thats write you plug in the point given at the beginning thank you!!

Answer 21

yw