найти производные первого и второго порядка [tex]1) y=ln(x+ \sqrt{48+ x^{2} } ) 2)y=cos( x^{20} )[/tex]. Created by daisywheell. matematika-ru

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daisywheell @daisywheell

July 2022 1 3 Report

найти производные первого и второго порядка
[tex]1) y=ln(x+ \sqrt{48+ x^{2} } )

2)y=cos( x^{20} )[/tex]

Answers & Comments

Artem112

Verified answer

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

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

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