Differentiate the following w.r.t. x :
`x^(sinx)+(sinx)^(cosx)`

Answer

To differentiate the function \( y = x^{\sin x} + (\sin x)^{\cos x} \) with respect to \( x \), we will break it down into two parts and differentiate each part separately.

### Step 1: Differentiate \( u = x^{\sin x} \)

1. **Let \( u = x^{\sin x} \)**
- Take the natural logarithm of both sides:
\[ \ln u = \sin x \cdot \ln x \]

2. **Differentiate both sides with respect to \( x \)** using implicit differentiation:
- Using the chain rule on the left side:
\[ \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\sin x \cdot \ln x) \]
- Apply the product rule on the right side:
\[ \frac{d}{dx}(\sin x \cdot \ln x) = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]

3. **Combine the results**:
\[ \frac{1}{u} \frac{du}{dx} = \cos x \cdot \ln x + \frac{\sin x}{x} \]
- Multiply both sides by \( u \):
\[ \frac{du}{dx} = u \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]
- Substitute back \( u = x^{\sin x} \):
\[ \frac{du}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

### Step 2: Differentiate \( v = (\sin x)^{\cos x} \)

1. **Let \( v = (\sin x)^{\cos x} \)**
- Take the natural logarithm:
\[ \ln v = \cos x \cdot \ln(\sin x) \]

2. **Differentiate both sides with respect to \( x \)**:
- Using the chain rule:
\[ \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(\cos x \cdot \ln(\sin x)) \]
- Apply the product rule on the right side:
\[ \frac{d}{dx}(\cos x \cdot \ln(\sin x)) = -\sin x \cdot \ln(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x \]

3. **Combine the results**:
\[ \frac{1}{v} \frac{dv}{dx} = -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \]
- Multiply both sides by \( v \):
\[ \frac{dv}{dx} = v \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]
- Substitute back \( v = (\sin x)^{\cos x} \):
\[ \frac{dv}{dx} = (\sin x)^{\cos x} \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]

### Step 3: Combine the derivatives

Now that we have \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
Substituting the expressions we derived:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]

### Final Result
Thus, the derivative of \( y = x^{\sin x} + (\sin x)^{\cos x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]