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Find an equation of the tangent line to the curve y=e^x/(1+x^2) at the point (1,1/2e)

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asked Aug 12, 2014 in CALCULUS by bananats Rookie

1 Answer

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The equation of the curve:

y = e^x/(1+x^2)

Take derivative with respect to x.

dy/dx =((1+x^2)d/dx(e^x)-e^xd/dx(1+x^2))/(1+x^2)^2

         =((1+x^2)(e^x)-e^x(2x))/(1+x^2)^2

         =e^x((1+x^2-2x)/(1+x^2)^2

         =e^x(1-x)^2/(1+x^2)^2

Slope at (1,1/2e)

m=( dy/dx)at (1,1/2e)

   =e^1(1-1)^2/(1+1^2)^2

   =0

Tangent equation:

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

y-(1/2e) =0(x-1)

y=1/2e

answered Aug 12, 2014 by bradely Mentor

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