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

Answer

To differentiate the function \( y = (\sin x)^{\cos x} + (\cos x)^{\sin x} \) with respect to \( x \), we will use logarithmic differentiation and the product rule. Let's break it down step by step.

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

### Step 2: Differentiate each term separately
We will differentiate \( u = (\sin x)^{\cos x} \) and \( v = (\cos x)^{\sin x} \) separately.

### Step 3: Differentiate \( u = (\sin x)^{\cos x} \)
Taking the natural logarithm of both sides:
\[ \log u = \cos x \cdot \log(\sin x) \]
Now differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\cos x \cdot \log(\sin x)) \]
Using the product rule on the right side:
\[ \frac{d}{dx}(\cos x \cdot \log(\sin x)) = -\sin x \cdot \log(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x \]
So we have:
\[ \frac{1}{u} \frac{du}{dx} = -\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x} \]
Multiplying through by \( u \):
\[ \frac{du}{dx} = u \left(-\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x}\right) \]
Substituting back \( u = (\sin x)^{\cos x} \):
\[ \frac{du}{dx} = (\sin x)^{\cos x} \left(-\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x}\right) \]

### Step 4: Differentiate \( v = (\cos x)^{\sin x} \)
Taking the natural logarithm of both sides:
\[ \log v = \sin x \cdot \log(\cos x) \]
Now differentiate both sides with respect to \( x \):
\[ \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(\sin x \cdot \log(\cos x)) \]
Using the product rule on the right side:
\[ \frac{d}{dx}(\sin x \cdot \log(\cos x)) = \cos x \cdot \log(\cos x) - \sin x \cdot \frac{1}{\cos x} \cdot (-\sin x) \]
So we have:
\[ \frac{1}{v} \frac{dv}{dx} = \cos x \cdot \log(\cos x) + \tan x \cdot \sin x \]
Multiplying through by \( v \):
\[ \frac{dv}{dx} = v \left(\cos x \cdot \log(\cos x) + \tan x \cdot \sin x\right) \]
Substituting back \( v = (\cos x)^{\sin x} \):
\[ \frac{dv}{dx} = (\cos x)^{\sin x} \left(\cos x \cdot \log(\cos x) + \tan x \cdot \sin x\right) \]

### Step 5: Combine the derivatives
Now, we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = (\sin x)^{\cos x} \left(-\sin x \cdot \log(\sin x) + \frac{\cos^2 x}{\sin x}\right) + (\cos x)^{\sin x} \left(\cos x \cdot \log(\cos x) + \tan x \cdot \sin x\right) \]

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