`(d)/(dx)[tan^(-1)((2+3tanx)/(3-2tanx))]=`

Options

A. 0
B. 1
C. `-1`
D. `3//2`

Answer

To solve the problem \((d)/(dx)[\tan^{-1}((2+3\tan x)/(3-2\tan x))]\), we will follow these steps:

### Step 1: Simplify the expression inside the arctangent
We start with the expression:
\[ y = \tan^{-1}\left(\frac{2 + 3\tan x}{3 - 2\tan x}\right) \]
To simplify this, we divide both the numerator and the denominator by 3:
\[ y = \tan^{-1}\left(\frac{\frac{2}{3} + \tan x}{1 - \frac{2}{3}\tan x}\right) \]

### Step 2: Substitute \(\frac{2}{3}\) with \(\tan \theta\)
Let \(\tan \theta = \frac{2}{3}\). Then we can rewrite our expression as:
\[ y = \tan^{-1}\left(\frac{\tan \theta + \tan x}{1 - \tan \theta \tan x}\right) \]

### Step 3: Use the tangent addition formula
Using the identity for the tangent of a sum, we have:
\[ \tan^{-1}\left(\frac{\tan \theta + \tan x}{1 - \tan \theta \tan x}\right) = \tan^{-1}(\tan(\theta + x)) \]
Thus, we can simplify \(y\) to:
\[ y = \theta + x \]

### Step 4: Replace \(\theta\) with \(\tan^{-1}\left(\frac{2}{3}\right)\)
Now substituting back for \(\theta\):
\[ y = \tan^{-1}\left(\frac{2}{3}\right) + x \]

### Step 5: Differentiate with respect to \(x\)
Now we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = 0 + 1 = 1 \]

### Final Answer
Thus, the derivative is:
\[ \frac{d}{dx}\left[\tan^{-1}\left(\frac{2 + 3\tan x}{3 - 2\tan x}\right)\right] = 1 \]