anonymous
  • anonymous
Please help, explanation would be greatly appreciated, picture included
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
welshfella
  • welshfella
First You need to find the derivative of y = arctan (x/3) this gives the slope of the tangent
welshfella
  • welshfella
can you do that?

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anonymous
  • anonymous
y'=x/(x^2+9) is what i got
welshfella
  • welshfella
i dont think thats correct let me check..
welshfella
  • welshfella
No its 3 / (x^2 + 9)
welshfella
  • welshfella
so at the origin (0,0) the slope is 3/9 = 1/3
anonymous
  • anonymous
do we use quotient rule for x/1?
welshfella
  • welshfella
No i just used the standard derivative for arctan (x / a)
welshfella
  • welshfella
y' = a / (a^2 + x^2)
anonymous
  • anonymous
i haven't learned that one yet i just know arctanu= u'/ (1+u^2)
welshfella
  • welshfella
- its basically the same thing
anonymous
  • anonymous
i understand where that comes from because u is over 1
welshfella
  • welshfella
yes
anonymous
  • anonymous
ok so what do we do once we have the correct derivative?
anonymous
  • anonymous
y'=3/(x^2+9)
welshfella
  • welshfella
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
anonymous
  • anonymous
can you please just give me the quick way you found the slope please
welshfella
  • welshfella
slope = 3 / (x^2 + 3^2) at the origin x = 0 so m = slope = 3 / (0 +9) = 1/3
welshfella
  • welshfella
i use the standard derivative of arctan (x/a) = a / (a^2 + x^2)
anonymous
  • anonymous
thats write you plug in the point given at the beginning thank you!!
welshfella
  • welshfella
yw

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