Найти производные первого и второго порядка [latex]1) y=ln(x+ \sqrt{48+ x^{2} } ) 2)y=cos( x^{20} )[/latex]

1) [latex]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}{2\sqrt{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}} \cdot\frac{\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}} [/latex] 2) [latex]y=\cos x^{20} \\\ y'=-\sin x^{20}\cdot(x^{20})'= -\sin x^{20}\cdot20x^{19}=-20x^{19}\sin x^{20} \\\ y''=-20\cdot((x^{19})'\cdot \sin x^{20}+x^{19}\cdot (\sin x^{20})')= \\\ =-20\cdot(19x^{18}\cdot \sin x^{20}+x^{19}\cdot \cos x^{20}\cdot(x^{20})')= \\\ =-20\cdot(19x^{18} \sin x^{20}+x^{19}\cdot \cos x^{20}\cdot20x^{19})= \\\ =-20\cdot(19x^{18} \sin x^{20}+20x^{38}\cos x^{20})= \\\ =-380x^{18} \sin x^{20}-400x^{38}\cos x^{20}[/latex]