The image asks for the first derivative (turunan pertama) of three different functions. Let's go through each one and find their derivatives.

8. $y = \frac{x^2}{2} + \sqrt{3x} - \frac{2}{x^2} + 2x$

Derivative:

1. $\frac{d}{dx} \left( \frac{x^2}{2} \right) = x$
2. $\frac{d}{dx} \left( \sqrt{3x} \right) = \frac{d}{dx} \left( (3x)^{1/2} \right) = \frac{1}{2} \cdot (3x)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x}}$
3. $\frac{d}{dx} \left( -\frac{2}{x^2} \right) = 4x^{-3}$
4. $\frac{d}{dx}(2x) = 2$

So, the derivative of the function is: $y' = x + \frac{3}{2\sqrt{3x}} + 4x^{-3} + 2$

9. $y = (3x - 2)(x^2 + 3)$

This requires the product rule: $\frac{d}{dx}(uv) = u'v + uv'$ where $u = 3x - 2$ and $v = x^2 + 3$.

1. $u' = 3$
2. $v' = 2x$

Now applying the product rule: $y' = (3)(x^2 + 3) + (3x - 2)(2x)$ Expanding both terms: $y' = 3x^2 + 9 + (6x^2 - 4x)$ $y' = 9x^2 - 4x + 9$

10. $y = \frac{x + 1}{\sqrt{x}}$

We can rewrite this as $y = (x + 1) \cdot x^{-1/2}$, and use the product rule again.

1. $u = x + 1$, so $u' = 1$
2. $v = x^{-1/2}$, so $v' = -\frac{1}{2}x^{-3/2}$

Applying the product rule: $y' = u'v + uv' = (1)(x^{-1/2}) + (x + 1)\left(-\frac{1}{2}x^{-3/2}\right)$ Expanding: $y' = x^{-1/2} - \frac{1}{2}(x + 1)x^{-3/2}$ Simplify further: $y' = \frac{1}{\sqrt{x}} - \frac{x + 1}{2x^{3/2}}$

Summary of the derivatives:

1. $y' = x + \frac{3}{2\sqrt{3x}} + 4x^{-3} + 2$
2. $y' = 9x^2 - 4x + 9$
3. $y' = \frac{1}{\sqrt{x}} - \frac{x + 1}{2x^{3/2}}$

Would you like any further clarification or detailed steps?

Here are 5 related questions to explore further:

1. How does the product rule apply to more complex products of functions?
2. Can you explain the chain rule for functions involving nested powers or roots?
3. How would you differentiate a quotient of two functions (using the quotient rule)?
4. What happens if you try to differentiate an implicit function?
5. How do you find the second derivative of these functions?

Tip: When dealing with roots and powers in derivatives, rewriting expressions with fractional exponents often simplifies the process.