Find:
\[
\frac{dy}{dx}, \quad \text{if} \quad y = x \tan x + \frac{\sqrt{x^2 + 1}}{2}
\]
Show Hint
When differentiating expressions involving trigonometric functions, don't forget to apply the product rule for terms like \( x \tan x \), and use the chain rule for composite functions like \( \frac{\sqrt{x^2 + 1}}{2} \).
We are given:
\[
y = x \tan x + \frac{\sqrt{x^2 + 1}}{2}
\]
Now, differentiate each term with respect to \( x \). The first term involves the product rule, and the second term involves the chain rule:
\[
\frac{dy}{dx} = \frac{d}{dx} \left( x \tan x \right) + \frac{d}{dx} \left( \frac{\sqrt{x^2 + 1}}{2} \right)
\]
Applying the product rule to \( x \tan x \) and using the chain rule for \( \frac{\sqrt{x^2 + 1}}{2} \), we get:
\[
\frac{dy}{dx} = \tan x + x \sec^2 x + \frac{x}{\sqrt{x^2 + 1}}
\]
