Question

найти производные первого и второго порядка
$1) y=ln(x+ sqrt{48+ x^{2} } ) 2)y=cos( x^{20} )$

Answer 1

1)
$y=ln(x+ sqrt{48+ x^{2} } ) \ y'= frac{1}{x+ sqrt{48+ x^2}} cdot(x+ sqrt{48+ x^2})'= \ =frac{1}{x+ sqrt{48+ x^2}} cdot(1+ frac{1}{2sqrt{48+ x^2}}cdot2x )= frac{1}{x+ sqrt{48+ x^2}} cdot(1+ frac{x}{sqrt{48+ x^2}} )= \ =frac{1}{x+sqrt{48+x^2}} cdotfrac{sqrt{48+x^2}+x}{sqrt{48+x^2}}=frac{1}{sqrt{48+ x^2}} =(x^2+48)^{- frac{1}{2} } \ y''=- frac{1}{2} (x^2+48)^{-frac{1}{2} -1}cdot(x^2-48)'= \ =-frac{1}{2}(x^2+48)^{- frac{3}{2}}cdot2x=-frac{x}{(x^2+48) sqrt{x^2+48}}$

2)
$y=cos x^{20} \ y'=-sin x^{20}cdot(x^{20})'= -sin x^{20}cdot20x^{19}=-20x^{19}sin x^{20} \ y''=-20cdot((x^{19})'cdot sin x^{20}+x^{19}cdot (sin x^{20})')= \ =-20cdot(19x^{18}cdot sin x^{20}+x^{19}cdot cos x^{20}cdot(x^{20})')= \ =-20cdot(19x^{18} sin x^{20}+x^{19}cdot cos x^{20}cdot20x^{19})= \ =-20cdot(19x^{18} sin x^{20}+20x^{38}cos x^{20})= \ =-380x^{18} sin x^{20}-400x^{38}cos x^{20}$