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

Dec 14, 2015

Derivative : $\frac{- 119 {x}^{2} + 98 x + 28}{3 {x}^{2} - 8 x + 4} ^ 2$

Explanation:

$h \left(x\right) = \left(\frac{2 x + 3}{x - 2}\right) \cdot \left(\frac{5 x - 1}{3 x - 2}\right)$

First, expand the function and combined like terms

$h \left(x\right) = \frac{10 {x}^{2} - 2 x + 15 x - 3}{3 {x}^{2} - 2 x - 6 x + 4}$

$= \frac{10 {x}^{2} + 13 x - 3}{3 {x}^{2} - 8 x + 4}$

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 $f \left(x\right) = 10 {x}^{2} + 13 x - 3$

$f ' \left(x\right) = 20 x + 13$

Let $g \left(x\right) = 3 {x}^{2} - 8 x + 4$

$g ' \left(x\right) = 6 x - 8$

$h ' \left(x\right) = \frac{\left(3 {x}^{2} - 8 x + 4\right) \left(20 x + 13\right) - \left(10 {x}^{2} + 13 x - 3\right) \left(6 x - 8\right)}{3 {x}^{2} - 8 x + 4} ^ 2$

$h ' \left(x\right) = \frac{\left(60 {x}^{3} - 121 {x}^{2} - 24 x + 52\right) - \left(60 {x}^{3} - 2 {x}^{2} - 122 x + 24\right)}{3 {x}^{2} - 8 x + 4} ^ 2$

$h ' \left(x\right) = \frac{60 {x}^{3} - 121 {x}^{2} - 24 x + 52 - 60 {x}^{3} + 2 {x}^{2} + 122 x - 24}{3 {x}^{2} - 8 x + 4} ^ 2$

$h ' \left(x\right) = \frac{- 119 {x}^{2} + 98 x + 28}{3 {x}^{2} - 8 x + 4} ^ 2$