Find the values of `(dy)/(dx)`, if `y = x^(tanx)+sqrt((x^(2)+1)/(2))`.

Explanation

To find the derivative \( \frac{dy}{dx} \) for the function

\[ y = x^{\tan x} + \sqrt{\frac{x^2 + 1}{2}}, \]

we will differentiate each part of the function separately.

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

Let \( u = x^{\tan x} \). To differentiate \( u \), we can use logarithmic differentiation.

Taking the natural logarithm of both sides:

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

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

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

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

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

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

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

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

Let \( v = \sqrt{\frac{x^2 + 1}{2}} \). We can simplify this to:

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

Now, differentiate \( v \):

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

### Step 3: Combine the derivatives

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

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

Substituting the expressions we found:

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

### 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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