If `y=x^(tanx)+sqrt((x^2+1)/2)` , find `(dy)/(dx)`

Explanation

To differentiate the function \( y = x^{\tan x} + \sqrt{\frac{x^2 + 1}{2}} \), we will follow these steps:

### Step 1: Differentiate \( x^{\tan x} \)

Let \( u = x^{\tan x} \).

To differentiate \( u \), we will take the natural logarithm of both sides:

\[ \ln u = \tan x \cdot \ln x \]

Now, differentiate both sides with respect to \( x \):

Using the chain rule on the left side and the product rule on the right side, we get:

\[ \frac{1}{u} \frac{du}{dx} = \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \]

Now, multiply both sides by \( u \):

\[ \frac{du}{dx} = u \left( \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \right) \]

Substituting back \( u = x^{\tan x} \):

\[ \frac{du}{dx} = x^{\tan x} \left( \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \right) \]

### Step 2: Differentiate \( \sqrt{\frac{x^2 + 1}{2}} \)

Now, we differentiate the second part of \( y \):

\[ v = \sqrt{\frac{x^2 + 1}{2}} = \frac{1}{\sqrt{2}} \sqrt{x^2 + 1} \]

Using the chain rule, we differentiate:

\[ \frac{dv}{dx} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{x}{\sqrt{2(x^2 + 1)}} \]

### Step 3: Combine the derivatives

Now, we can combine the derivatives to find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]

Substituting the expressions we found:

\[ \frac{dy}{dx} = x^{\tan x} \left( \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \right) + \frac{x}{\sqrt{2(x^2 + 1)}} \]

### Final Answer

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = x^{\tan x} \left( \sec^2 x \cdot \ln x + \frac{\tan x}{x} \right) + \frac{x}{\sqrt{2(x^2 + 1)}} \]

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