# Write an equation for the tangent to y=(4x)/(x^2+1) at the point (1,2).

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### 1 Answer

first we have to find the derivative of `y=(4x)/(x^2+1)`

`y'=((x^2+1)(4)-4x(2x))/(x^2+1)^2 = (4x^2+4-8x^2)/(x^2+1)^2=-(4x^2-4)/(x^2+1)^2`

`y'(1)=(4-4)/(1+1)^2=0`

So the slope is 0, so the tangent is a horizontal line.

y=2