Question

If y= 3x^2 + 2x/ 2x^2 - 3 calculate dy/dx help me plz

Answer 1

Option A:

`$(dy)/(dx)= (-2(2x^2+9x+3) )/((2x^2-3)^2)$`

Solution:

Given
`$y=(3x^2+2x)/(2x^2-3) $`

To calculate
`(dy)/(dx)`:

`$(dy)/(dx)=(d)/(dx)((3x^2+2x)/(2x^2-3)) $`

Using differential rule:

`$(d)/(dx)((u)/(v) ) =(v(du)/(dx)-u(dv)/(dx) )/(v^2)$`

Here,
`u=3x^2+2x` and
`v=2x^2-3`

`$(d)/(dx)((3x^2+2x)/(2x^2-3)) =((2x^2-3)(d)/(dx)(3x^2+2x)-(3x^2+2x)(d)/(dx)(2x^2-3) )/((2x^2-3)^2)$` – – – – (1)

Now, using another differential rule:
`(d)/(dx) x^n=nx^(n-1)` and
`(d)/(dx) (a)=0`, a=constant

`$(d)/(dx)(3x^2+2x) =6x+2` – – – – (2)

`$(d)/(dx)(2x^2-3) =4x` – – – – (3)

Substitute (2) and (3) in (1), we get

`$(d)/(dx)((3x^2+2x)/(2x^2-3)) =((2x^2-3)(6x+2)-(3x^2+2x)(4x) )/((2x^2-3)^2)$`

Now, simplifying the above equation

`$=((12x^3+4x^2-18x-6)-(12x^3+8x^2) )/((2x^2-3)^2)$`

`$=(12x^3+4x^2-18x-6-12x^3-8x^2 )/((2x^2-3)^2)$`

`$=(-4x^2-18x-6 )/((2x^2-3)^2)$`

Take –2 as common factor in numerator of the fraction.

`$=(-2(2x^2+9x+3) )/((2x^2-3)^2)$`

`$(dy)/(dx)= (-2(2x^2+9x+3) )/((2x^2-3)^2)$`