Question

Найти производные функции (y'): 1) y = 5^4/x^2 2) y = ln (x+4/x^2+4) 3) y = 5^sinx 4) y = 8^arccos(2x)

Answer 1

$1)y`= (5^{ \frac{4}{ x^{2} }})` =[(a ^{u} )`=a ^{u}\cdot u`]= 5^{ \frac{4}{ x^{2} }}\cdot( \frac{4}{ x^{2} })` =5^{ \frac{4}{ x^{2} }}\cdot( 4\cdot x^{-2} })` = \\ =[x ^{ \alpha }= \alpha x^{ \alpha -1}]= 5^{ \frac{4}{ x^{2} }}\cdot 4\cdot(-2)\cdot x^{-3} }= -8\cdot 5^{ \frac{4}{ x^{2} }}\cdot \frac{1}{ x^{3} }$
$2) y` = ln `( \frac{x+4}{ x^{2} +4})=[(lnu)`= \frac{u`}{u}]= \frac{1}{ \frac{x+4}{ x^{2} +4} }\cdot(\frac{x+4}{ x^{2} +4})`= \\ =[( \frac{u}{v})`= \frac{u`v-uv`}{v ^{2} }]= \frac{x^{2} +4}{ x +4} \cdot\frac{(x+4)`(x^{2} +4) -(x+4)(x^{2} +4)` }{ (x^{2} +4) ^{2} }= \\ = \frac{1}{ x +4} \cdot\frac{1\cdot(x^{2} +4) -(x+4)\cdot 2x }{ (x^{2} +4) }= \frac{- x^{2} -8x+4}{(x+4)( x^{2} +4)}$
$3) y` = (5^{sinx})`=[(a ^{u})`=a ^{u}\cdot lna\cdot u`]=5^{sinx}\cdot ln5\cdot(sinx)`= \\ =5^{sinx}\cdot ln5\cdot cosx$
$4) y` = (8^{arccos2x})`=[(a ^{u})`=a ^{u}\cdot lna\cdot u`]=8^{arccos2x}\cdot ln8\cdot (arccos2x)`$
$=[(arccosu)`=- \frac{1}{ \sqrt{1-u ^{2} } }\cdot u`]=8^{arccos2x}\cdot ln8\cdot (- \frac{1}{ \sqrt{1-(2x) ^{2} } }\cdot (2x)`)= \\ 8^{arccos2x}\cdot ln8\cdot (- \frac{1}{ \sqrt{1-4x ^{2} } })\cdot 2$