To help you with exercise 3 from the image, I'll extract and analyze the text related to it.

Exercise 3:

The problem states:

Determină $\frac{dy}{dx}$ știind că y este o funcție implicită a variabilei $x$, definită de ecuația: $y - 2x - \arcsin\left(\frac{y}{2}\right) = 0.$

Translation: Determine $\frac{dy}{dx}$, knowing that $y$ is an implicit function of the variable $x$, defined by the equation: $y - 2x - \arcsin\left(\frac{y}{2}\right) = 0.$

Solution:

We will use implicit differentiation to solve this.

The equation is: $y - 2x - \arcsin\left(\frac{y}{2}\right) = 0.$

1. Differentiate both sides of the equation with respect to $x$. Remember that $y$ is a function of $x$, so we will use the chain rule where necessary.

$\frac{d}{dx} \left( y - 2x - \arcsin\left(\frac{y}{2}\right) \right) = \frac{d}{dx}(0).$

2. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$, the derivative of $-2x$ is $-2$, and the derivative of $\arcsin\left(\frac{y}{2}\right)$ involves the chain rule:

$\frac{d}{dx} \arcsin\left(\frac{y}{2}\right) = \frac{1}{\sqrt{1 - \left( \frac{y}{2} \right)^2}} \cdot \frac{1}{2} \cdot \frac{dy}{dx}.$

Thus, the derivative of the whole expression is:

$\frac{dy}{dx} - 2 - \frac{1}{\sqrt{1 - \left(\frac{y}{2}\right)^2}} \cdot \frac{1}{2} \cdot \frac{dy}{dx} = 0.$

3. Simplify the equation to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} \left( 1 - \frac{1}{2\sqrt{1 - \left( \frac{y}{2} \right)^2}} \right) = 2.$

4. Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{2}{1 - \frac{1}{2\sqrt{1 - \left( \frac{y}{2} \right)^2}}}.$

This is the derivative $\frac{dy}{dx}$ for the implicit function.

Let me know if you'd like more clarification on any step or if you have additional questions! Here are some related questions for further exploration:

1. How do we apply the chain rule in implicit differentiation?
2. What is the derivative of $\arcsin(x)$?
3. How would the process change if $y$ was an explicit function of $x$?
4. Can we simplify the result of $\frac{dy}{dx}$ further?
5. How do we interpret the solution in terms of the geometry of the curve?

Tip: Always check if the expression inside functions like arcsine or logarithm stays within the defined domain when solving real-world problems.