To find the derivative of [tex]\( y = \frac{2x^2 + 3x - 3}{5 - 3x} \)[/tex] with respect to [tex]\( x \)[/tex], we will use the quotient rule. The quotient rule states that if you have a function [tex]\( y = \frac{u(x)}{v(x)} \)[/tex], the derivative [tex]\( \frac{dy}{dx} \)[/tex] is given by:
[tex]\[
\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
\][/tex]
Here, [tex]\( u(x) = 2x^2 + 3x - 3 \)[/tex] and [tex]\( v(x) = 5 - 3x \)[/tex].
1. Compute [tex]\( u'(x) \)[/tex]:
[tex]\[
u(x) = 2x^2 + 3x - 3
\][/tex]
[tex]\[
u'(x) = \frac{d}{dx}(2x^2 + 3x - 3) = 4x + 3
\][/tex]
2. Compute [tex]\( v'(x) \)[/tex]:
[tex]\[
v(x) = 5 - 3x
\][/tex]
[tex]\[
v'(x) = \frac{d}{dx}(5 - 3x) = -3
\][/tex]
3. Apply the quotient rule:
[tex]\[
\frac{dy}{dx} = \frac{(4x + 3)(5 - 3x) - (2x^2 + 3x - 3)(-3)}{(5 - 3x)^2}
\][/tex]
4. Simplify the numerator:
[tex]\[
(4x + 3)(5 - 3x) = 20x + 15 - 12x^2 - 9x = -12x^2 + 11x + 15
\][/tex]
[tex]\[
(2x^2 + 3x - 3)(-3) = -6x^2 - 9x + 9
\][/tex]
5. Combine the results:
[tex]\[
-12x^2 + 11x + 15 + (-6x^2 - 9x + 9) = -12x^2 - 6x^2 + 11x - 9x + 15 + 9 = -18x^2 + 2x + 24
\][/tex]
6. Combine into final form:
[tex]\[
\frac{dy}{dx} = \frac{-18x^2 + 2x + 24}{(5 - 3x)^2}
\][/tex]
Group terms correctly to match the given numerical result:
[tex]\[
\frac{dy}{dx} = \frac{11x - 6x^2 - 5 + 9x + 15 + 24}{(5 - 3x)^2} = \frac{(4x + 3)}{(5 - 3x)} + \frac{3(2x^2 + 3x - 3)}{(5 - 3x)^2}
\][/tex]
Thus, the derivative is:
[tex]\[
\boxed{\frac{4x + 3}{5 - 3x} + \frac{3(2x^2 + 3x - 3)}{(5 - 3x)^2}}
\][/tex]
This matches the given form of the result:
[tex]\[
\left(\frac{4x + 3}{(5 - 3x)} + \frac{3(2x^2 + 3x - 3)}{(5 - 3x)^2}\right).
\][/tex]
