Find the derivative of
`2/(x+1)-x^(2)/(3x-1)`.

Answer

To find the derivative of the function \( f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1} \), we will apply the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then its derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

### Step 1: Differentiate the first term \( \frac{2}{x+1} \)

Let \( u = 2 \) and \( v = x + 1 \).

- The derivative of \( u \) is \( \frac{du}{dx} = 0 \).
- The derivative of \( v \) is \( \frac{dv}{dx} = 1 \).

Using the quotient rule:

\[ \frac{d}{dx}\left(\frac{2}{x+1}\right) = \frac{(x+1) \cdot 0 - 2 \cdot 1}{(x+1)^2} = \frac{-2}{(x+1)^2} \]

### Step 2: Differentiate the second term \( -\frac{x^2}{3x-1} \)

Let \( u = x^2 \) and \( v = 3x - 1 \).

- The derivative of \( u \) is \( \frac{du}{dx} = 2x \).
- The derivative of \( v \) is \( \frac{dv}{dx} = 3 \).

Using the quotient rule:

\[ \frac{d}{dx}\left(-\frac{x^2}{3x-1}\right) = -\frac{(3x-1)(2x) - (x^2)(3)}{(3x-1)^2} \]

Simplifying this gives:

\[ = -\frac{(6x^2 - 2x - 3x^2)}{(3x-1)^2} = -\frac{3x^2 - 2x}{(3x-1)^2} \]

### Step 3: Combine the derivatives

Now we combine the derivatives from Step 1 and Step 2:

\[ f'(x) = \frac{-2}{(x+1)^2} - \left(-\frac{3x^2 - 2x}{(3x-1)^2}\right) \]

This simplifies to:

\[ f'(x) = \frac{-2}{(x+1)^2} + \frac{3x^2 - 2x}{(3x-1)^2} \]

### Final Answer

Thus, the derivative of the given function \( f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1} \) is:

\[ f'(x) = \frac{-2}{(x+1)^2} + \frac{3x^2 - 2x}{(3x-1)^2} \]