Find `(dy)/(dx)`, when:
`y=x^(sinx)+(sinx)^(cosx)`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = x^{\sin x} + (\sin x)^{\cos x}\), we will differentiate each part of the function separately and then combine the results.

### Step-by-Step Solution:

1. **Identify the two parts of the function:**
\[ y = u + v \]
where \(u = x^{\sin x}\) and \(v = (\sin x)^{\cos x}\).

2. **Differentiate \(u = x^{\sin x}\):**
- Take the natural logarithm of both sides:
\[ \ln u = \sin x \cdot \ln x \]
- Differentiate both sides with respect to \(x\):
\[ \frac{1}{u} \frac{du}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]
- Multiply through by \(u\):
\[ \frac{du}{dx} = u \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]
- Substitute back for \(u\):
\[ \frac{du}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

3. **Differentiate \(v = (\sin x)^{\cos x}\):**
- Take the natural logarithm of both sides:
\[ \ln v = \cos x \cdot \ln(\sin x) \]
- Differentiate both sides with respect to \(x\):
\[ \frac{1}{v} \frac{dv}{dx} = -\sin x \cdot \ln(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x \]
- Simplifying gives:
\[ \frac{dv}{dx} = v \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]
- Substitute back for \(v\):
\[ \frac{dv}{dx} = (\sin x)^{\cos x} \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]

4. **Combine the derivatives:**
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]

### Final Answer:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]