What is the slope of the tangent to the curve y = 2x/(x2 + 1) at (0, 0)?

The correct option is 2.

To find the slope of the tangent to the curve at a given point, we need to calculate the first derivative of the function, which is represented by $\frac{dy}{dx}$, and evaluate it at the given point $(0,0)$.

The equation of the curve is:

$$y=\frac{2x}{x^2+1}$$

We can find the derivative using the quotient rule of differentiation. The quotient rule states that if we have a function of the form $y=\frac{u}{v}$, its derivative is given by:

$$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$

For our function, let:
$u=2x$ which gives $\frac{du}{dx}=2$
$v=x^2+1$ which gives $\frac{dv}{dx}=2x$

Now, substitute these derivatives into the quotient rule formula:

$$\frac{dy}{dx}=\frac{(x^2+1)\cdot 2-\left(2x\right)\cdot \left(2x\right)}{{(x^2+1)}^2}$$

Simplify the expression in the numerator:

$$\frac{dy}{dx}=\frac{2x^2+2-4x^2}{{(x^2+1)}^2}$$

$$\frac{dy}{dx}=\frac{2-2x^2}{{(x^2+1)}^2}$$

To find the slope of the tangent at the point $(0,0)$, evaluate the derivative at $x=0$:

$${|\frac{dy}{dx}}_{x=0}=\frac{2-2{\left(0\right)}^2}{{(0^2+1)}^2}$$

$${|\frac{dy}{dx}}_{x=0}=\frac{2}{1}=2$$

Thus, the slope of the tangent to the curve at the point $(0,0)$ is indeed $2$.