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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)

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  1. myininaya
    • 5 years ago
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    find y' first

  2. myininaya
    • 5 years ago
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    to find slope

  3. anonymous
    • 5 years ago
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    how would i do that?

  4. myininaya
    • 5 years ago
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    use chain rule

  5. anonymous
    • 5 years ago
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    with arctan?

  6. anonymous
    • 5 years ago
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    what does arctan change to?

  7. myininaya
    • 5 years ago
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    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)

  8. myininaya
    • 5 years ago
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    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

  9. myininaya
    • 5 years ago
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    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

  10. myininaya
    • 5 years ago
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    when y=pi/4 so the tangent line goes through point (2,pi/4)

  11. myininaya
    • 5 years ago
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    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

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