Question

Знайдіть похідну 1)f(х)=4х²+12х+9/х²-16
2)f(х)=х+1/х²+5​

Answer 1

Ответ:

1) $\boxed{f(x)' = - \dfrac{12x^{2} + 146x + 192 }{(x^{2} - 16)^{2}}}$

2) $\boxed{f(x)' = \dfrac{ 5 - 2x - x^{2} }{(x^{2} + 5)^{2}}}$

Объяснение:

1) $f(x) = \dfrac{4x^{2} + 12x + 9}{x^{2} - 16}$

$f(x)' = \left ( \dfrac{4x^{2} + 12x + 9}{x^{2} - 16} \right)' = \dfrac{(4x^{2} + 12x + 9)' (x^{2} - 16) - (4x^{2} + 12x + 9) (x^{2} - 16)'}{(x^{2} - 16)^{2}}=$$= \dfrac{(8x + 12 ) (x^{2} - 16) - 2x(4x^{2} + 12x + 9) }{(x^{2} - 16)^{2}}= \dfrac{8x^{3} -128x + 12x^{2} - 192 - (8x^{3} + 24x^{2} + 18x) }{(x^{2} - 16)^{2}}=$$= \dfrac{8x^{3} - 128x + 12x^{2} - 192 - 8x^{3} - 24x^{2} - 18x }{(x^{2} - 16)^{2}} =\dfrac{ - 12x^{2} - 192 - 146x }{(x^{2} - 16)^{2}} =$

$= - \dfrac{12x^{2} + 146x + 192 }{(x^{2} - 16)^{2}};$

2) $f(x) = \dfrac{x + 1}{x^{2} + 5}$

$f(x)' = \left( \dfrac{x + 1}{x^{2} + 5} \right)' = \dfrac{(x + 1)'(x^{2} + 5) - (x + 1)(x^{2} + 5)'}{(x^{2} + 5)^{2}} = \dfrac{1(x^{2} + 5) - 2x(x + 1)}{(x^{2} + 5)^{2}} =$$= \dfrac{(x^{2} + 5) - 2x(x + 1)}{(x^{2} + 5)^{2}} = \dfrac{x^{2} + 5 - (2x^{2} + 2x)}{(x^{2} + 5)^{2}} = \dfrac{x^{2} + 5 - 2x^{2} - 2x}{(x^{2} + 5)^{2}} =$

$= \dfrac{ 5 - 2x - x^{2} }{(x^{2} + 5)^{2}};$