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

Answer

To differentiate the function \( y = x^{\cos x} + (\cos x)^x \) with respect to \( x \), we will follow the steps outlined below:

### Step 1: Rewrite the function in exponential form
We start by rewriting the function using the property of logarithms:
\[ y = x^{\cos x} + (\cos x)^x \]
We can express \( x^{\cos x} \) and \( (\cos x)^x \) in terms of the exponential function:
\[ y = e^{\cos x \ln x} + e^{x \ln(\cos x)} \]

### Step 2: Differentiate using the chain rule
Now we will differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{\cos x \ln x}) + \frac{d}{dx}(e^{x \ln(\cos x)}) \]

Using the chain rule, we have:
\[ \frac{dy}{dx} = e^{\cos x \ln x} \cdot \frac{d}{dx}(\cos x \ln x) + e^{x \ln(\cos x)} \cdot \frac{d}{dx}(x \ln(\cos x)) \]

### Step 3: Differentiate the first term
For the first term \( \frac{d}{dx}(\cos x \ln x) \), we apply the product rule:
\[ \frac{d}{dx}(\cos x \ln x) = \frac{d}{dx}(\cos x) \cdot \ln x + \cos x \cdot \frac{d}{dx}(\ln x) \]
Calculating the derivatives:
\[ \frac{d}{dx}(\cos x) = -\sin x, \quad \frac{d}{dx}(\ln x) = \frac{1}{x} \]
Thus,
\[ \frac{d}{dx}(\cos x \ln x) = -\sin x \ln x + \frac{\cos x}{x} \]

### Step 4: Differentiate the second term
Now for the second term \( \frac{d}{dx}(x \ln(\cos x)) \):
Using the product rule again:
\[ \frac{d}{dx}(x \ln(\cos x)) = \frac{d}{dx}(x) \cdot \ln(\cos x) + x \cdot \frac{d}{dx}(\ln(\cos x)) \]
Calculating the derivatives:
\[ \frac{d}{dx}(x) = 1, \quad \frac{d}{dx}(\ln(\cos x)) = -\tan x \]
Thus,
\[ \frac{d}{dx}(x \ln(\cos x)) = \ln(\cos x) - x \tan x \]

### Step 5: Combine results
Now we can combine everything:
\[ \frac{dy}{dx} = e^{\cos x \ln x} \left(-\sin x \ln x + \frac{\cos x}{x}\right) + e^{x \ln(\cos x)} \left(\ln(\cos x) - x \tan x\right) \]

### Step 6: Substitute back the original expressions
Substituting back \( e^{\cos x \ln x} = x^{\cos x} \) and \( e^{x \ln(\cos x)} = (\cos x)^x \):
\[ \frac{dy}{dx} = x^{\cos x} \left(-\sin x \ln x + \frac{\cos x}{x}\right) + (\cos x)^x \left(\ln(\cos x) - x \tan x\right) \]

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