Step 1:

We notice that the given function is a quotient of two other functions. To find the derivative of this, we can use the quotient rule. First, let’s label the two functions as:

[latex]\large{f(x) = x^2-3x+1}[/latex] and [latex]\large{g(x) = 2x^2+5}[/latex]

To use the quotient rule, we also need the derivative of these functions. We can easily find the derivatives as:

[latex]\large{f'(x) = 2x-3}[/latex] and [latex]\large{g'(x) = 4x}[/latex]

Step 2:

The quotient rule formula states that:

[latex]\large{\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f'(x) – f(x)g'(x)}{g(x)^2}}[/latex]

Thus, substituting into the quotient rule formula, we have:

[latex]\large{\frac{d}{dx}(\frac{x^2-3x+1}{2x^2+5})}[/latex]

[latex]\large{ = \frac{(2x^2+5)(2x-3) \:- \:(x^2-3x+1)(4x)}{(2x^2+5)^2}}[/latex]

Step 3:

Simplifying, we have:

[latex]\Large{ = \frac{4x^3-6x^2+10x-15-4x^3+12x^2-4x}{(2x^2+5)^2}}[/latex]

[latex]\Large{ = \frac{6x^2+6x-15}{(2x^2+5)^2}}[/latex]

Thus, this is the final answer for the derivative. So:

[latex]\Large{\frac{dy}{dx} = \frac{6x^2+6x-15}{(2x^2+5)^2}}[/latex]

Solution Complete!