If `y=(tanx+cotx)/(tanx-cotx)," then: "(dy)/(dx)=`

Options

A. `2sec2x*tan2x`
B. `sec2x*tan2x`
C. `-sec2x*tan2x`
D. `-2sec2x*tan2x`

Answer

To find the derivative of the function \( y = \frac{\tan x + \cot x}{\tan x - \cot x} \), we will follow these steps:

### Step 1: Rewrite the function
We know that \( \cot x = \frac{1}{\tan x} \). Therefore, we can rewrite \( y \) as:
\[ y = \frac{\tan x + \frac{1}{\tan x}}{\tan x - \frac{1}{\tan x}} \]

### Step 2: Simplify the expression
To simplify the expression, we can multiply the numerator and the denominator by \( \tan x \):
\[ y = \frac{\tan^2 x + 1}{\tan^2 x - 1} \]

### Step 3: Use the identity for secant
We know that \( \tan^2 x + 1 = \sec^2 x \) and \( \tan^2 x - 1 = \sec^2 x - 2 \). Thus, we can rewrite \( y \) as:
\[ y = \frac{\sec^2 x}{\tan^2 x - 1} \]

### Step 4: Rewrite using cosine
Using the identity \( \tan^2 x = \frac{1 - \cos 2x}{\sin^2 2x} \), we can express \( y \) in terms of cosine:
\[ y = \frac{1 + \tan^2 x}{\tan^2 x - 1} = \frac{1 + \tan^2 x}{\sec^2 x - 2} \]

### Step 5: Differentiate \( y \)
Now we will differentiate \( y \) with respect to \( x \):
Using the quotient rule:
\[ \frac{dy}{dx} = \frac{(v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx})}{v^2} \]
where \( u = \sec^2 x \) and \( v = \tan^2 x - 1 \).

### Step 6: Calculate derivatives
Calculating the derivatives:
- \( \frac{du}{dx} = 2 \sec^2 x \tan x \)
- \( \frac{dv}{dx} = 2 \tan x \sec^2 x \)

### Step 7: Substitute back into the quotient rule
Substituting back into the quotient rule:
\[ \frac{dy}{dx} = \frac{(\tan^2 x - 1)(2 \sec^2 x \tan x) - (\sec^2 x)(2 \tan x \sec^2 x)}{(\tan^2 x - 1)^2} \]

### Step 8: Simplify the expression
After simplification, we find:
\[ \frac{dy}{dx} = -2 \sec 2x \tan 2x \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -2 \sec 2x \tan 2x \]

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