Differentiate y = (1)/(1+2 x) with respect to x.

We are given the function:

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

Recall the Pythagorean trigonometric identity: $1 + \tan^2 x = \sec^2 x$.
Substitute this into the expression for $y$:

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

Since $\sec x = \frac{1}{\cos x}$, it follows that $\frac{1}{\sec x} = \cos x$. Therefore, $\frac{1}{\sec^2 x} = \cos^2 x$.

$$y = \cos^2 x$$

Let $u = \cos x$. Then $y = u^2$.
The derivative of $y$ with respect to $u$ is $\frac{dy}{du} = 2u$.
The derivative of $u$ with respect to $x$ is $\frac{du}{dx} = -\sin x$.
Applying the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$$y' = 2u \cdot (-\sin x)$$

Substitute $u = \cos x$ back into the equation:

$$y' = 2(\cos x)(-\sin x)$$ $$y' = -2 \sin x \cos x$$

Recall the double angle identity for sine: $\sin(2x) = 2 \sin x \cos x$.
Substitute this into the expression for $y'$:

$$y' = -\sin(2x)$$

The final answer is $-\sin(2x)$.