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 given function

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

we will use the properties of the inverse tangent function and the chain rule for differentiation.

### Step 1: Simplifying the Function
Using the identity for the tangent inverse function, we can rewrite the first term:

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

For the second term, we can also apply the identity:

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

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

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

Notice that \(-\tan^{-1}(x) + \tan^{-1}(x)\) cancels out, leading to:

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

### Step 2: Differentiating \(y\)
Now we differentiate \(y\) with respect to \(x\):

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

The derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1 + x^2}\). Therefore, applying the chain rule:

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

Since \(\tan^{-1}(2)\) is a constant, its derivative is 0:

\[ \frac{d}{dx}(\tan^{-1}(2)) = 0. \]

### Step 3: Final Result
Combining these results, we have:

\[ \frac{dy}{dx} = \frac{5}{1 + 25x^2} + 0 = \frac{5}{1 + 25x^2}. \]

Thus, the final answer is:

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

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