differentiate-y-3x-2-2-2-6x-2-x-3-1-

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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