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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




Slope at (1,1/2e)

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



Tangent equation:


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


answered Aug 12, 2014 by bradely Mentor

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