`y=e^(sinx)+(tanx)^(x)`

Answer

To find the derivative of the function \( y = e^{\sin x} + (\tan x)^x \), we will differentiate it step by step.

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

### Step 2: Differentiate using the sum rule
The derivative of a sum is the sum of the derivatives. Thus, we can write:
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{\sin x}) + \frac{d}{dx}((\tan x)^x) \]

### Step 3: Differentiate \( e^{\sin x} \)
Using the chain rule:
\[ \frac{d}{dx}(e^{\sin x}) = e^{\sin x} \cdot \frac{d}{dx}(\sin x) = e^{\sin x} \cdot \cos x \]

### Step 4: Differentiate \( (\tan x)^x \)
To differentiate \( (\tan x)^x \), we can use logarithmic differentiation. Let:
\[ y_2 = (\tan x)^x \]
Taking the natural logarithm of both sides:
\[ \ln y_2 = x \ln(\tan x) \]
Now differentiate both sides with respect to \( x \):
\[ \frac{1}{y_2} \frac{dy_2}{dx} = \ln(\tan x) + x \cdot \frac{1}{\tan x} \cdot \sec^2 x \]
Thus:
\[ \frac{dy_2}{dx} = y_2 \left( \ln(\tan x) + x \cdot \frac{\sec^2 x}{\tan x} \right) \]
Substituting back \( y_2 = (\tan x)^x \):
\[ \frac{dy_2}{dx} = (\tan x)^x \left( \ln(\tan x) + x \cdot \frac{\sec^2 x}{\tan x} \right) \]

### Step 5: Combine the derivatives
Now, substituting back into the derivative of \( y \):
\[ \frac{dy}{dx} = e^{\sin x} \cos x + (\tan x)^x \left( \ln(\tan x) + x \cdot \frac{\sec^2 x}{\tan x} \right) \]

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

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