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

Answer

To differentiate the function \( y = x^{\sin x} + \cos^x \) with respect to \( x \), we will use the properties of logarithms and the chain rule. Here’s a step-by-step solution:

### Step 1: Rewrite the Function
We start with the function:
\[ y = x^{\sin x} + \cos^x \]

### Step 2: Differentiate Each Term
We will differentiate each term separately.

#### For the first term \( x^{\sin x} \):
Using the property \( a^b = e^{b \ln a} \):
\[ x^{\sin x} = e^{\sin x \ln x} \]
Now, differentiate using the chain rule:
\[ \frac{dy_1}{dx} = e^{\sin x \ln x} \cdot \frac{d}{dx}(\sin x \ln x) \]
Using the product rule on \( \sin x \ln x \):
\[ \frac{d}{dx}(\sin x \ln x) = \cos x \ln x + \sin x \cdot \frac{1}{x} \]
Thus,
\[ \frac{dy_1}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) \]

#### For the second term \( \cos^x \):
Using the property \( a^b = e^{b \ln a} \):
\[ \cos^x = e^{x \ln(\cos x)} \]
Now, differentiate using the chain rule:
\[ \frac{dy_2}{dx} = e^{x \ln(\cos x)} \cdot \frac{d}{dx}(x \ln(\cos x)) \]
Using the product rule on \( x \ln(\cos x) \):
\[ \frac{d}{dx}(x \ln(\cos x)) = \ln(\cos x) + x \cdot \frac{1}{\cos x} \cdot (-\sin x) \]
Thus,
\[ \frac{dy_2}{dx} = \cos^x \left( \ln(\cos x) - x \tan x \right) \]

### Step 3: Combine the Derivatives
Now, we can combine the derivatives of both terms:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) + \cos^x \left( \ln(\cos x) - x \tan x \right) \]

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

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