Differentiate the following w.r.t. x :
`(tanx)^(cotx)+x^(tanx),0ltxltpi/4`

Answer

To differentiate the function \( y = (\tan x)^{\cot x} + x^{\tan x} \) with respect to \( x \), we will use the properties of logarithms and the chain rule. Let's break it down step by step.

### Step 1: Differentiate \( y = (\tan x)^{\cot x} \)

Let \( p = (\tan x)^{\cot x} \).

To differentiate \( p \), we will take the natural logarithm of both sides:

\[ \ln p = \cot x \cdot \ln(\tan x) \]

Now, differentiate both sides with respect to \( x \):

\[ \frac{1}{p} \frac{dp}{dx} = \frac{d}{dx}(\cot x \cdot \ln(\tan x)) \]

Using the product rule on the right side:

\[ \frac{d}{dx}(\cot x) \cdot \ln(\tan x) + \cot x \cdot \frac{d}{dx}(\ln(\tan x)) \]

Now, we know:

\[ \frac{d}{dx}(\cot x) = -\csc^2 x \]
\[ \frac{d}{dx}(\ln(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \]

So, substituting these derivatives back, we have:

\[ \frac{1}{p} \frac{dp}{dx} = -\csc^2 x \cdot \ln(\tan x) + \cot x \cdot \frac{\sec^2 x}{\tan x} \]

Multiplying through by \( p \):

\[ \frac{dp}{dx} = p \left( -\csc^2 x \cdot \ln(\tan x) + \cot x \cdot \frac{\sec^2 x}{\tan x} \right) \]

Substituting back \( p = (\tan x)^{\cot x} \):

\[ \frac{dp}{dx} = (\tan x)^{\cot x} \left( -\csc^2 x \cdot \ln(\tan x) + \cot x \cdot \frac{\sec^2 x}{\tan x} \right) \]

### Step 2: Differentiate \( y = x^{\tan x} \)

Let \( q = x^{\tan x} \).

Taking the natural logarithm:

\[ \ln q = \tan x \cdot \ln x \]

Differentiating both sides:

\[ \frac{1}{q} \frac{dq}{dx} = \frac{d}{dx}(\tan x \cdot \ln x) \]

Using the product rule:

\[ \frac{d}{dx}(\tan x) \cdot \ln x + \tan x \cdot \frac{d}{dx}(\ln x) \]

We know:

\[ \frac{d}{dx}(\tan x) = \sec^2 x \]
\[ \frac{d}{dx}(\ln x) = \frac{1}{x} \]

So substituting these back:

\[ \frac{1}{q} \frac{dq}{dx} = \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \]

Multiplying through by \( q \):

\[ \frac{dq}{dx} = x^{\tan x} \left( \sec^2 x \cdot \ln x + \frac{\tan x}{x} \right) \]

### Step 3: Combine the results

Now, we have:

\[ \frac{dy}{dx} = \frac{dp}{dx} + \frac{dq}{dx} \]

Substituting the expressions we found for \( \frac{dp}{dx} \) and \( \frac{dq}{dx} \):

\[ \frac{dy}{dx} = (\tan x)^{\cot x} \left( -\csc^2 x \cdot \ln(\tan x) + \cot x \cdot \frac{\sec^2 x}{\tan x} \right) + x^{\tan x} \left( \sec^2 x \cdot \ln x + \frac{\tan x}{x} \right) \]

This gives us the final derivative of the function.