Find `(dy)/(dx)` if `y=tan^(-1)(4x)/(1+5x^2)+tan^(-1)(2+3x)/(3-2x)`

Explanation

To find \(\frac{dy}{dx}\) for the function

\[ y = \tan^{-1}\left(\frac{4x}{1 + 5x^2}\right) + \tan^{-1}\left(\frac{2 + 3x}{3 - 2x}\right), \]

we can follow these steps:

### Step 1: Rewrite the Function
The first term can be rewritten using the identity for the tangent of a difference:

\[ \tan^{-1}\left(\frac{4x}{1 + 5x^2}\right) = \tan^{-1}(5x) - \tan^{-1}(x). \]

The second term can also be rewritten:

\[ \tan^{-1}\left(\frac{2 + 3x}{3 - 2x}\right) = \tan^{-1}\left(\frac{2/3 + x}{1 - \frac{2}{3}x}\right) = \tan^{-1}\left(\frac{2}{3}\right) + \tan^{-1}(x). \]

Thus, we can express \(y\) as:

\[ y = \tan^{-1}(5x) - \tan^{-1}(x) + \tan^{-1}\left(\frac{2}{3}\right) + \tan^{-1}(x). \]

### Step 2: Simplify the Expression
Notice that \(-\tan^{-1}(x) + \tan^{-1}(x)\) cancels out, leaving us with:

\[ y = \tan^{-1}(5x) + \tan^{-1}\left(\frac{2}{3}\right). \]

### Step 3: Differentiate the Function
Now, we differentiate \(y\) with respect to \(x\):

\[ \frac{dy}{dx} = \frac{d}{dx} \left(\tan^{-1}(5x)\right) + \frac{d}{dx}\left(\tan^{-1}\left(\frac{2}{3}\right)\right). \]

Since \(\tan^{-1}\left(\frac{2}{3}\right)\) is a constant, its derivative is \(0\). Therefore, we only need to differentiate \(\tan^{-1}(5x)\):

Using the derivative formula for \(\tan^{-1}(u)\):

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}, \]

where \(u = 5x\). Thus, we have:

\[ \frac{du}{dx} = 5. \]

Now substituting \(u\) back into the derivative:

\[ \frac{dy}{dx} = \frac{1}{1 + (5x)^2} \cdot 5 = \frac{5}{1 + 25x^2}. \]

### Final Answer
Thus, the final result is:

\[ \frac{dy}{dx} = \frac{5}{1 + 25x^2}. \]