Question

How do you differentiate `f(x)= [(2x+3)/(x-2)][(5x-1)/(3x-2)]` using the chain rule.?

Answer 1

Derivative : `(-119x^2 +98x+28)/(3x^2-8x+4)^2`

Explanation:

`h(x) = ((2x+3)/(x-2))*((5x-1)/(3x-2))`

First, expand the function and combined like terms

`h(x) = (10x^2-2x+15x-3)/(3x^2 -2x-6x+4)`

`=(10x^2 +13x-3)/(3x^2 -8x+4)`

We can differentiate this using quotient rule
`h(x)= f(x)/g(x) ; " " h'(x) = (f'(x) g(x) - g'(x)f(x))/(g(x))^2`

Let `f(x) = 10x^2 +13x -3`

`f'(x) = 20x+13`

Let `g(x) = 3x^2 -8x+4`

`g'(x) = 6x-8`

`h'(x) = ((3x^2-8x+4)(20x+13) -(10x^2+13x-3)(6x-8))/(3x^2-8x+4)^2`

`h'(x) = ((60x^3-121x^2-24x+52)-(60x^3- 2x^2-122x+24))/(3x^2-8x+4)^2`

`h'(x) = (60x^3 - 121x^2 -24x+52-60x^3+2x^2+122x-24)/(3x^2-8x+4)^2`

`h'(x) = (-119x^2 +98x+28)/(3x^2-8x+4)^2`