155/BBDF. problem / Math - Solved Math Problems

MR-155 / 24. problem

Determine the first derivative of the function below:

$y = \frac{2 x^{2} + 3 x}{4 x - 6}$

$y = \frac{2 x^{2} + 3 x}{4 x - 6}$

$y = \frac{u}{v} = \frac{2 x^{2} + 3 x}{4 x - 6}$

$u = 2 x^{2} + 3 x; u^{'} = 4 x + 3$

$v = 4 x - 6 \Rightarrow v^{'} = 4$

$y^{'} = \frac{u^{'} v - u v^{'}}{v^{2}} = \frac{(4 x + 3) \cdot (4 x - 6) - (2 x^{2} + 3 x) \cdot 4}{{(4 x - 6)}^{2}}$

$y^{'} = \frac{16 x^{2} - 24 x + 12 x - 18 - 8 x^{2} - 12 x}{{[2 (2 x - 3)]}^{2}}$

$y^{'} = \frac{8 x^{2} - 24 x - 18}{2^{2} {(2 x - 3)}^{2}}$

$y^{'} = \frac{2 \cdot (4 x^{2} - 12 x - 9)}{4 {(2 x - 3)}^{2}}$

$y^{'} = \frac{4 x^{2} - 12 x - 9}{2 \cdot {(2 x - 3)}^{2}}$

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