Tangent line to a sum of transcendental functions

First find the derivative: $y' = e^x + \frac{1}{1+x^2}$, since $\frac{d}{dx}[e^x] = e^x$ and $\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$. Evaluate the slope at $x=0$: $y'(0) = e^0 + \frac{1}{1+0^2} = 1 + 1 = 2$.

Next find the point on the curve at $x=0$: $y(0) = e^0 + \arctan(0) = 1 + 0 = 1$, so the point is $(0,1)$. Using the point-slope form $y - 1 = 2(x - 0)$, the tangent line is $y = 2x + 1$.

The slope was found by evaluating the derivative at x equals zero. Since y prime equals e to the x plus one over one plus x squared, plugging in zero gives one plus one, which equals two. So the answer is two.