Answer: [tex]f^{\prime}(x) = \boxed{\quad}[/tex]

Given:

[tex]\[ f(x) = 2x^3 - x^2 - 5x + 3 \][/tex]

1. For the first term [tex]\( 2x^3 \)[/tex]:

[tex]\[ \frac{d}{dx}(2x^3) = 2 \times 3x^{3-1} = 6x^2 \][/tex]

2. For the second term [tex]\( -x^2 \)[/tex]:

[tex]\[ \frac{d}{dx}(-x^2) = -1 \times 2x^{2-1} = -2x \][/tex]

3. For the third term [tex]\( -5x \)[/tex]:

[tex]\[ \frac{d}{dx}(-5x) = -5 \][/tex]

4. For the constant term [tex]\( 3 \)[/tex]:

[tex]\[ \frac{d}{dx}(3) = 0 \][/tex]

Now, combine these derivatives:

[tex]\[ f'(x) = 6x^2 - 2x - 5 \][/tex]

Thus, the derivative of the function [tex]\( f(x) = 2x^3 - x^2 - 5x + 3 \)[/tex] is:

[tex]\[ f'(x) = 6x^2 - 2x - 5 \][/tex]