If `y=tan^(-1)((6+5 tanx)/(5- 6 tanx))`, then `(dy)/(dx)=`

Options

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

Answer

To find the derivative of the function \( y = \tan^{-1}\left(\frac{6 + 5 \tan x}{5 - 6 \tan x}\right) \), we will use the chain rule and the derivative of the inverse tangent function. Let's solve it step by step.

### Step 1: Differentiate the function
We start with the function:
\[ y = \tan^{-1}\left(\frac{6 + 5 \tan x}{5 - 6 \tan x}\right) \]

Using the chain rule, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{6 + 5 \tan x}{5 - 6 \tan x}\right)^2} \cdot \frac{d}{dx}\left(\frac{6 + 5 \tan x}{5 - 6 \tan x}\right) \]

### Step 2: Differentiate the inner function
Let \( u = \frac{6 + 5 \tan x}{5 - 6 \tan x} \). We need to find \( \frac{du}{dx} \).

Using the quotient rule:
\[ \frac{du}{dx} = \frac{(5 - 6 \tan x)(5 \sec^2 x) - (6 + 5 \tan x)(-6 \sec^2 x)}{(5 - 6 \tan x)^2} \]

### Step 3: Simplify \( \frac{du}{dx} \)
Now we simplify:
\[ \frac{du}{dx} = \frac{(5 - 6 \tan x)(5 \sec^2 x) + (6 + 5 \tan x)(6 \sec^2 x)}{(5 - 6 \tan x)^2} \]
\[ = \frac{(25 \sec^2 x - 30 \tan x \sec^2 x + 36 \sec^2 x + 30 \tan x \sec^2 x)}{(5 - 6 \tan x)^2} \]
\[ = \frac{(61 \sec^2 x)}{(5 - 6 \tan x)^2} \]

### Step 4: Substitute back into the derivative
Now we substitute \( \frac{du}{dx} \) back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{6 + 5 \tan x}{5 - 6 \tan x}\right)^2} \cdot \frac{61 \sec^2 x}{(5 - 6 \tan x)^2} \]

### Step 5: Simplify the expression
To simplify \( 1 + \left(\frac{6 + 5 \tan x}{5 - 6 \tan x}\right)^2 \):
\[ 1 + \left(\frac{6 + 5 \tan x}{5 - 6 \tan x}\right)^2 = \frac{(5 - 6 \tan x)^2 + (6 + 5 \tan x)^2}{(5 - 6 \tan x)^2} \]

Calculating the numerator:
\[ (5 - 6 \tan x)^2 + (6 + 5 \tan x)^2 = (25 - 60 \tan x + 36 \tan^2 x) + (36 + 60 \tan x + 25 \tan^2 x) \]
\[ = 61 + 61 \tan^2 x \]

Thus, we have:
\[ \frac{dy}{dx} = \frac{61 \sec^2 x}{(5 - 6 \tan x)^2} \cdot \frac{(5 - 6 \tan x)^2}{61 + 61 \tan^2 x} \]
\[ = \frac{61 \sec^2 x}{61 + 61 \tan^2 x} \]
\[ = \frac{\sec^2 x}{1 + \tan^2 x} \]
Since \( 1 + \tan^2 x = \sec^2 x \), we finally get:
\[ \frac{dy}{dx} = 1 \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 1 \]