`y=(sinx)^(cosx)+(cosx)^(sinx)`

Answer

To find the derivative of the function \( y = (\sin x)^{\cos x} + (\cos x)^{\sin x} \), we will differentiate each term separately using logarithmic differentiation.

### Step-by-Step Solution:

1. **Define the Function**:
\[ y = (\sin x)^{\cos x} + (\cos x)^{\sin x} \]

2. **Differentiate the First Term** \( y_1 = (\sin x)^{\cos x} \):
- Take the natural logarithm of both sides:
\[ \log y_1 = \cos x \cdot \log(\sin x) \]
- Differentiate both sides with respect to \( x \):
\[ \frac{1}{y_1} \frac{dy_1}{dx} = -\sin x \cdot \log(\sin x) + \cos x \cdot \frac{\cos x}{\sin x} \]
- Rearranging gives:
\[ \frac{dy_1}{dx} = y_1 \left( -\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]
- Substitute \( y_1 \):
\[ \frac{dy_1}{dx} = (\sin x)^{\cos x} \left( -\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]

3. **Differentiate the Second Term** \( y_2 = (\cos x)^{\sin x} \):
- Take the natural logarithm of both sides:
\[ \log y_2 = \sin x \cdot \log(\cos x) \]
- Differentiate both sides with respect to \( x \):
\[ \frac{1}{y_2} \frac{dy_2}{dx} = \cos x \cdot \log(\cos x) - \sin x \cdot \tan x \]
- Rearranging gives:
\[ \frac{dy_2}{dx} = y_2 \left( \cos x \cdot \log(\cos x) - \sin x \cdot \tan x \right) \]
- Substitute \( y_2 \):
\[ \frac{dy_2}{dx} = (\cos x)^{\sin x} \left( \cos x \cdot \log(\cos x) - \sin x \cdot \tan x \right) \]

4. **Combine the Derivatives**:
- The total derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]
- Substituting the expressions for \( \frac{dy_1}{dx} \) and \( \frac{dy_2}{dx} \):
\[ \frac{dy}{dx} = (\sin x)^{\cos x} \left( -\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x} \right) + (\cos x)^{\sin x} \left( \cos x \cdot \log(\cos x) - \sin x \cdot \tan x \right) \]

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