Differentiate the following w.r.t. x :
`(sinx-cosx)^(sinx-cosx),pi/4ltxlt(3pi)/4`

Answer

To differentiate the function \( y = (\sin x - \cos x)^{(\sin x - \cos x)} \) with respect to \( x \), we will use logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \ln y = \ln\left((\sin x - \cos x)^{(\sin x - \cos x)}\right) \]

### 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:
\[ \ln y = (\sin x - \cos x) \ln(\sin x - \cos x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). Using implicit differentiation on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left[(\sin x - \cos x) \ln(\sin x - \cos x)\right] \]

### Step 4: Apply the product rule
Let \( u = \sin x - \cos x \) and \( v = \ln(\sin x - \cos x) \). Then, using the product rule \( \frac{d}{dx}(uv) = u'v + uv' \):
- \( u' = \cos x + \sin x \)
- \( v = \ln u \) implies \( v' = \frac{1}{u} u' = \frac{1}{\sin x - \cos x} (\cos x + \sin x) \)

Thus,
\[ \frac{d}{dx} \left[(\sin x - \cos x) \ln(\sin x - \cos x)\right] = (\cos x + \sin x) \ln(\sin x - \cos x) + (\sin x - \cos x) \cdot \frac{1}{\sin x - \cos x} (\cos x + \sin x) \]

### Step 5: Simplify the right-hand side
The right-hand side simplifies to:
\[ (\cos x + \sin x) \ln(\sin x - \cos x) + (\cos x + \sin x) = (\cos x + \sin x) \left( \ln(\sin x - \cos x) + 1 \right) \]

### Step 6: Substitute back into the equation
Now substituting back into our differentiated equation:
\[ \frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) \left( \ln(\sin x - \cos x) + 1 \right) \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot (\cos x + \sin x) \left( \ln(\sin x - \cos x) + 1 \right) \]

### Step 8: Substitute \( y \) back
Recall that \( y = (\sin x - \cos x)^{(\sin x - \cos x)} \):
\[ \frac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)} \cdot (\cos x + \sin x) \left( \ln(\sin x - \cos x) + 1 \right) \]

### Final Answer
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) \left( \ln(\sin x - \cos x) + 1 \right) \]