Question

Знайти похідну функції g(x)=x2-4x/x+1 у точці х0=-2 ?

Answer 1

$g(x) = \frac{ {x}^{2} - 4x }{x + 1} \\ g'(x) = \frac{( {x}^{2} - 4x)' \times (x +1 ) - (x + 1)' \times (x {}^{2} - 4x) }{(x + 1) {}^{2} } = \\ \frac{(2x - 4)(x + 1) - ( {x }^{2} - 4x) }{ {x}^{2} + 2x + 1 } = \\ \frac{2 {x}^{2} + 2x - 4x - 4 - {x}^{2} + 4x }{ {x}^{2} + 2x + 1} = \\ \frac{ {x}^{2} +2 x - 4}{ {x}^{2} + 2x + 1} = \frac{ {x}^{2} + 2x - 4 }{(x + 1) {}^{2} } \\ x_{0} = - 2 \\ g'(x) = \frac{( - 2) {}^{2} + 2 \times ( - 2) - 4}{( - 2 + 1) {}^{2} } = \\ \frac{4 - 4 - 4}{1} = - 4$