differentiate y=(((3x^2 +2)^2 (√(6x+2)))/(x^3 +1))


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Question Number 129026 by oustmuchiya@gmail.com last updated on 12/Jan/21

$d i f f e r e n t i a t e y = \frac{{(3 x^{2} + 2)}^{2} \sqrt{6 x + 2}}{x^{3} + 1}$
	Answered by MJS_new last updated on 12/Jan/21

	$y = \frac{u v}{w}$ $y^{'} = \frac{{(u v)}^{'} w - w^{'} u v}{w^{2}} = \frac{(u^{'} v + v^{'} u) w - w^{'} u v}{w^{2}} =$ $= \frac{u^{'} v + v^{'} u}{w} - \frac{w^{'} u v}{w^{2}}$ $u = {(3 x^{2} + 2)}^{2}$ $v = \sqrt{6 x + 2}$ $w = x^{3} + 1$ $now {it}^{'} s easy . I get$ $y^{'} = \frac{3 \sqrt{2} (3 x^{2} + 2) (9 x^{5} + 2 x^{4} - 10 x^{3} + 23 x^{2} + 8 x + 2)}{2 {(x + 1)}^{2} {(x^{2} - x + 1)}^{2} \sqrt{3 x + 1}}$
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