$$ \frac{\differentialD y}{\differentialD x} $$$$ \left(x\text{ tan(}0.5\cdot x\right) $$

It seems like you're asking to differentiate the following expression:

\[
y = x \cdot \tan\left(\frac{x}{2}\right)
\]

Let's differentiate this with respect to $x$. We will use the product rule, which states that:

\[
\frac{d}{dx}(u \cdot v) = u'v + uv'
\]

### Steps:

1. Let $ u = x $ and $ v = \tan\left(\frac{x}{2}\right) $.
2. First, find the derivative of $u = x$:
   \[
   u' = \frac{d}{dx}(x) = 1
   \]
3. Now, find the derivative of $ v = \tan\left(\frac{x}{2}\right) $:
   \[
   v' = \frac{d}{dx}\left(\tan\left(\frac{x}{2}\right)\right)
   \]
   Using the chain rule:
   \[
   v' = \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2}
   \]

4. Apply the product rule:
   \[
   \frac{dy}{dx} = u'v + uv' = 1 \cdot \tan\left(\frac{x}{2}\right) + x \cdot \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2}
   \]

Thus, the derivative is:

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

### Final answer:
\[
\frac{dy}{dx} = \tan\left(\frac{x}{2}\right) + \frac{x}{2} \cdot \sec^2\left(\frac{x}{2}\right)
\]

Let me know if you'd like further clarification on any steps!

### Related questions:
1. What is the product rule and how is it applied?
2. How do you differentiate the tangent function?
3. What is the chain rule and when is it used in differentiation?
4. Can you explain how to differentiate trigonometric functions?
5. How does the secant function behave as a derivative of tangent?

### Tip:
When differentiating products of functions, always remember the product rule. Keep track of which function you're differentiating in each term to avoid mistakes.

Differentiate the following expression: $$ \frac{d}{dx} \left(x \cdot \tan\left(\frac{x}{2}\right)\right) $$

Learn how to differentiate the expression x * tan(x/2) step-by-step using the product rule and chain rule. This problem covers essential calculus concepts including trigonometric differentiation and proper application of the product rule. Suitable for advanced high school or college-level students.

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