If `y = (tan x)^(cot x) " then " (dy)/(dx)=` ?

Options

A. `cot x.(tan x)^(cot x -1). Sec^(2) x`
B. `- (tan x)^(cot x). "cosec"^(2)c`
C. `(tanx)^(cot x)."cosec"^(2)x (1 - log tan x)`
D. none of these

Explanation

To solve the problem \( y = (\tan x)^{\cot x} \) and find \( \frac{dy}{dx} \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the expression:
\[ \ln y = \ln((\tan x)^{\cot x}) \]

### Step 2: Apply the power rule of logarithms
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can rewrite the equation:
\[ \ln y = \cot x \cdot \ln(\tan x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). Remember to use the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cot x \cdot \ln(\tan x)) \]

Using the product rule:
\[ \frac{d}{dx}(\cot x \cdot \ln(\tan x)) = \cot x \cdot \frac{d}{dx}(\ln(\tan x)) + \ln(\tan x) \cdot \frac{d}{dx}(\cot x) \]

### Step 4: Differentiate \( \ln(\tan x) \) and \( \cot x \)
Now we need to find the derivatives:
- The derivative of \( \ln(\tan x) \) is:
\[ \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \]
- The derivative of \( \cot x \) is:
\[ -\csc^2 x \]

### Step 5: Substitute the derivatives back into the equation
Substituting these derivatives back into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = \cot x \cdot \frac{\sec^2 x}{\tan x} - \csc^2 x \cdot \ln(\tan x) \]

### Step 6: Simplify the expression
We can simplify \( \cot x \cdot \frac{\sec^2 x}{\tan x} \):
\[ \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x} \quad \Rightarrow \quad \frac{\sec^2 x}{\tan x} = \frac{1}{\sin x} \]
Thus,
\[ \cot x \cdot \frac{\sec^2 x}{\tan x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \frac{\cos x}{\sin^2 x} \]

### Step 7: Final expression for \( \frac{dy}{dx} \)
Now we can write:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{\cos x}{\sin^2 x} - \csc^2 x \cdot \ln(\tan x) \]
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \frac{\cos x}{\sin^2 x} - \csc^2 x \cdot \ln(\tan x) \right) \]
Substituting back \( y = (\tan x)^{\cot x} \):
\[ \frac{dy}{dx} = (\tan x)^{\cot x} \left( \frac{\cos x}{\sin^2 x} - \csc^2 x \cdot \ln(\tan x) \right) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = (\tan x)^{\cot x} \left( \frac{\cos x}{\sin^2 x} - \csc^2 x \cdot \ln(\tan x) \right) \]