Question

Нужно найти производную: $x=\frac{sqrt(x+2)*(3-x)^4}{(x+1)^5}$

Answer 1

Ответ:

$y = \frac{ \sqrt{x + 2} \times {(3 - x)}^{4} }{ {(x + 1)}^{5} }$

найдем с помощью логарифмической производной:

$y' = ( ln(y))' \times y$

$( ln(y))' = ( ln( \frac{ {(x + 2)}^{ \frac{1}{2} } \times {(3 - x)}^{4} }{ {(x + 1)}^{5} } )' = \\ = ( ln( {(x + 2)}^{ \frac{1}{2} } + ln( {(3 - x)}^{4} ) - ln( {(x + 1)}^{5} ) )' = \\ = \frac{1}{2(x + 2)} - \frac{4}{(3 - x)} - \frac{5}{x + 1} = \\ = \frac{(x + 1)(3 - x) - 8(x + 2)(x + 1) - 10(x + 2)(3 - x)}{2(x + 2)(x + 1)(3 - x)} = \\ = \frac{ {x}^{2} - 32x - 73}{2(x + 1)(x + 2)(3 - x)}$

$y' = \frac{ \sqrt{x + 2} \times {(3 - x)}^{4} }{ {(x + 1)}^{5} } \times \frac{ {x}^{2} - 32x - 73 }{2(x + 1)(x + 2)(3 - x)} = \\ = \frac{( {x}^{2} - 32x - 73) {(3 - x)}^{3} }{ 2{(x + 1)}^{6} \sqrt{x + 2} }$