Question

Find an equation of the tangent line at the point where x=2 for the function y=arctan(x/2)

Answer 1

find y' first

Answer 2

to find slope

Answer 3

how would i do that?

Answer 4

use chain rule

Answer 5

with arctan?

Answer 6

what does arctan change to?

Answer 7

let u=x/2 so we have y=arctan(u)
so dy/dx=1/(u^2+1) * du/dx
=1/(u^2+1) * 1/2
=1/([x/2]^2+1) * 1/2
=1/[x^2/4+1) * 1/2
=4/(x^2+4) * 1/2
=2/{x^2+4)

Answer 8

now to find the slope at x=2 we just plug in in to y'
y'(x=2)=2/(2^2+4)=2/8=1/4
so the slope is 1/4
now all you have to do is find the y-intercept
in write the tangent line in this form y=mx+b where m=1/4

Answer 9

we know a point on this line (2, arctan(2/2))
arctan(2/2)=arctan(1)
if arctan(1)=y
then
tany=1
so what makes the above true
reflect back to your unit circle
tany is the same as siny/cosy
so you want to find on your unit cirlce where siny is the same as cosy
when are they same

Answer 10

when y=pi/4
so the tangent line goes through point (2,pi/4)

Answer 11

y=mx+b
y=x/4+b
pi/4=2/4+b
(pi-2)/4=b
so the equation of the tangent line that runs through (2,pi/4) is
y=x/4+pi/4-1/2