Question

Найти значение второй производной f''(x) в точке х0=1 если f(x)=5-х^3 / х^2

Answer 1

$f`(x)=( \frac{5- x^{3} }{ x^{2} } )`=[( \frac{u}{v})`= \frac{u`v-uv`}{v ^{2} }]= \frac{(5- x^{3}) ` x^{2} -(5- x^{3}) (x^{2} )`}{( x^{2} ) ^{2} }= \\ \\ \frac{-3 x^{2} \cdot x^{2} -(5- x^{3})\cdot 2x }{ x^{4} }= \frac{-3 x^{4}-10x+2x^{4} }{ x^{4} }= \frac{- x^{4}-10x }{x ^{4} }$
$f``(x)=(\frac{- x^{4}-10x }{x ^{4} } )`= \frac{(-x ^{4}-10x)`\cdot x^{4} -(- x^{4}-10x)\cdot( x^{4})` }{( x^{4})^{2} }= \frac{-4 x^{7}-10 x^{4} +4 x^{7}+40x ^{4} }{ x^{8} } = \\ \\ = \frac{30}{x ^{4} }$

f``(1)=30


2 способ


$f`(x)= (\frac{5}{ x^{2} }-x)`=- \frac{10}{ x^{3} }-1 \\ \\ f``(x)=(- \frac{10}{ x^{3} }-1 )`= \frac{30}{ x^{4} }-0$
f``(1)=30