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

$1)y`= (5^{ frac{4}{ x^{2} }})` =[(a ^{u} )`=a ^{u}cdot u`]= 5^{ frac{4}{ x^{2} }}cdot( frac{4}{ x^{2} })` =5^{ frac{4}{ x^{2} }}cdot( 4cdot x^{-2} })` = \ =[x ^{ alpha }= alpha x^{ alpha -1}]= 5^{ frac{4}{ x^{2} }}cdot 4cdot(-2)cdot x^{-3} }= -8cdot 5^{ frac{4}{ x^{2} }}cdot frac{1}{ x^{3} }$
$2) y` = ln `( frac{x+4}{ x^{2} +4})=[(lnu)`= frac{u`}{u}]= frac{1}{ frac{x+4}{ x^{2} +4} }cdot(frac{x+4}{ x^{2} +4})`= \ =[( frac{u}{v})`= frac{u`v-uv`}{v ^{2} }]= frac{x^{2} +4}{ x +4} cdotfrac{(x+4)`(x^{2} +4) -(x+4)(x^{2} +4)` }{ (x^{2} +4) ^{2} }= \ = frac{1}{ x +4} cdotfrac{1cdot(x^{2} +4) -(x+4)cdot 2x }{ (x^{2} +4) }= frac{- x^{2} -8x+4}{(x+4)( x^{2} +4)}$
$3) y` = (5^{sinx})`=[(a ^{u})`=a ^{u}cdot lnacdot u`]=5^{sinx}cdot ln5cdot(sinx)`= \ =5^{sinx}cdot ln5cdot cosx$
$4) y` = (8^{arccos2x})`=[(a ^{u})`=a ^{u}cdot lnacdot u`]=8^{arccos2x}cdot ln8cdot (arccos2x)`$
$=[(arccosu)`=- frac{1}{ sqrt{1-u ^{2} } }cdot u`]=8^{arccos2x}cdot ln8cdot (- frac{1}{ sqrt{1-(2x) ^{2} } }cdot (2x)`)= \ 8^{arccos2x}cdot ln8cdot (- frac{1}{ sqrt{1-4x ^{2} } })cdot 2$