Question

Знайти похідну функцій.
$y = \frac{4x ^{2} + 8x + 3 }{2x + 4}$
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Answer 1

$y = \frac{4 {x}^{2} + 8x + 3 }{2x + 4}$

$y' = \frac{(4 {x}^{2} + 8x + 3)'(2x + 4) - (4 {x}^{2} + 8x + 3)(2x + 4) ' }{(2x + 4) {}^{2} } = \frac{(8x + 8)(2x + 4) - 2(4 {x}^{2} + 8x + 3) }{(2x + 4) {}^{2} } = \frac{16(x + 1)(x + 2) - 2(4 {x}^{2} + 8x + 3)}{4(x + 2) {}^{2} } = \frac{(8( {x}^{2} + 3x + 2) - 4 {x}^{2} - 8x - 3) }{2(x + 2) {}^{2} } = \frac{8 {x}^{2} + 24x + 16 - 4 {x}^{2} - 8x - 3 }{2(x + 2) {}^{2} } = \frac{4 {x}^{2} + 16x + 13 }{2( {x}^{2} + 4x + 4) } = \frac{4 {x}^{2} + 16x + 13}{2 {x}^{2} + 8x + 8}$