How do you find the derivative of y= [(2x+3)/(x-2)][(5x-1)/(3x-2)] using the chain rule?

$y ' = \frac{- 119 {x}^{2} + 98 x + 28}{3 {x}^{2} - 8 x + 4} ^ 2$

Explanation:

From the given equation.

$y = \left[\frac{2 x + 3}{x - 2}\right] \left[\frac{5 x - 1}{3 x - 2}\right]$

Simplify the right side of the equation by multiplying the rational expressions first

$y = \left[\frac{2 x + 3}{x - 2}\right] \left[\frac{5 x - 1}{3 x - 2}\right]$

$y = \left[\frac{10 {x}^{2} + 13 x - 3}{3 {x}^{2} - 8 x + 4}\right]$

Find the derivative by using the formula $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

Let $u = 10 {x}^{2} + 13 x - 3$

Let $v = 3 {x}^{2} - 8 x + 4$

Use now the formula

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

$y ' = \frac{d}{\mathrm{dx}} \left(\frac{10 {x}^{2} + 13 x - 3}{3 {x}^{2} - 8 x + 4}\right)$

$y ' = \frac{\left(3 {x}^{2} - 8 x + 4\right) \cdot \frac{d}{\mathrm{dx}} \left(10 {x}^{2} + 13 x - 3\right) - \left(10 {x}^{2} + 13 x - 3\right) \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 8 x + 4\right)}{3 {x}^{2} - 8 x + 4} ^ 2$

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

Simplify the numerator

$y ' = \frac{60 {x}^{3} - 160 {x}^{2} + 80 x + 39 {x}^{2} - 104 x + 52 - \left(60 {x}^{3} + 78 {x}^{2} - 18 x - 80 {x}^{2} - 104 x + 24\right)}{3 {x}^{2} - 8 x + 4} ^ 2$

$y ' = \frac{60 {x}^{3} - 160 {x}^{2} + 80 x + 39 {x}^{2} - 104 x + 52 - 60 {x}^{3} - 78 {x}^{2} + 18 x + 80 {x}^{2} + 104 x - 24}{3 {x}^{2} - 8 x + 4} ^ 2$

$y ' = \frac{- 119 {x}^{2} + 98 x + 28}{3 {x}^{2} - 8 x + 4} ^ 2$

God bless....I hope the explanation is useful.