Option A:
[tex]$\frac{dy}{dx}= \frac{-2(2x^2+9x+3) }{(2x^2-3)^2}$[/tex]
Solution:
Given [tex]$y=\frac{3x^2+2x}{2x^2-3} $[/tex]
To calculate [tex]\frac{dy}{dx}[/tex]:
[tex]$\frac{dy}{dx}=\frac{d}{dx}(\frac{3x^2+2x}{2x^2-3}) $[/tex]
Using differential rule:
[tex]$\frac{d}{dx}(\frac{u}{v} ) =\frac{v\frac{du}{dx}-u\frac{dv}{dx} }{v^2}$[/tex]
Here, [tex]u=3x^2+2x[/tex] and [tex]v=2x^2-3[/tex]
[tex]$\frac{d}{dx}(\frac{3x^2+2x}{2x^2-3}) =\frac{(2x^2-3)\frac{d}{dx}(3x^2+2x)-(3x^2+2x)\frac{d}{dx}(2x^2-3) }{(2x^2-3)^2}$[/tex] – – – – (1)
Now, using another differential rule: [tex]\frac{d}{dx} x^n=nx^{n-1}[/tex] and [tex]\frac{d}{dx} (a)=0[/tex], a=constant
[tex]$\frac{d}{dx}(3x^2+2x) =6x+2[/tex] – – – – (2)
[tex]$\frac{d}{dx}(2x^2-3) =4x[/tex] – – – – (3)
Substitute (2) and (3) in (1), we get
[tex]$\frac{d}{dx}(\frac{3x^2+2x}{2x^2-3}) =\frac{(2x^2-3)(6x+2)-(3x^2+2x)(4x) }{(2x^2-3)^2}$[/tex]
Now, simplifying the above equation
[tex]$=\frac{(12x^3+4x^2-18x-6)-(12x^3+8x^2) }{(2x^2-3)^2}$[/tex]
[tex]$=\frac{12x^3+4x^2-18x-6-12x^3-8x^2 }{(2x^2-3)^2}$[/tex]
[tex]$=\frac{-4x^2-18x-6 }{(2x^2-3)^2}$[/tex]
Take –2 as common factor in numerator of the fraction.
[tex]$=\frac{-2(2x^2+9x+3) }{(2x^2-3)^2}$[/tex]
[tex]$\frac{dy}{dx}= \frac{-2(2x^2+9x+3) }{(2x^2-3)^2}$[/tex]
