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 use the properties of logarithmic differentiation and the product rule. Here's the step-by-step solution:

### Step 1: Define the function
Let:
\[ y = x^{\sin x} + (\sin x)^{\cos x} \]

### Step 2: Differentiate the first term \( x^{\sin x} \)
Let:
\[ u = x^{\sin x} \]
Taking the natural logarithm of both sides:
\[ \ln u = \sin x \cdot \ln x \]
Now, differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]
Multiplying through by \( u \):
\[ \frac{du}{dx} = u \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]
Substituting back for \( u \):
\[ \frac{du}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

### Step 3: Differentiate the second term \( (\sin x)^{\cos x} \)
Let:
\[ v = (\sin x)^{\cos x} \]
Taking the natural logarithm of both sides:
\[ \ln v = \cos x \cdot \ln(\sin x) \]
Now, differentiate both sides with respect to \( x \):
\[ \frac{1}{v} \frac{dv}{dx} = -\sin x \cdot \frac{1}{\sin x} \cdot \cos x + \cos x \cdot \frac{\cos x}{\sin x} \]
This simplifies to:
\[ \frac{1}{v} \frac{dv}{dx} = -\cos x + \frac{\cos^2 x}{\sin x} \]
Multiplying through by \( v \):
\[ \frac{dv}{dx} = v \left( -\cos x + \frac{\cos^2 x}{\sin x} \right) \]
Substituting back for \( v \):
\[ \frac{dv}{dx} = (\sin x)^{\cos x} \left( -\cos x + \frac{\cos^2 x}{\sin x} \right) \]

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

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