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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what does arctan change to?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0let 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)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0now 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 yintercept in write the tangent line in this form y=mx+b where m=1/4

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we 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

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0when y=pi/4 so the tangent line goes through point (2,pi/4)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0y=mx+b y=x/4+b pi/4=2/4+b (pi2)/4=b so the equation of the tangent line that runs through (2,pi/4) is y=x/4+pi/41/2
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