найти 2 производную(х+cos 1/x^2)''. Created by 556641. matematika-ru

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nelle987

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$(x+\cos(1/x^2))''=x''+(\cos(1/x^2))''=(\cos x^{-2})''=(2x^{-3}\sin x^{-2})'=\\=-6x^{-4}\sin x^{-2}+2x^{-3}\cdot(-2x^{-3})\cos x^{-2}=-\dfrac{6x^2\sin x^{-2}+4\cos x^{-2}}{x^6}$

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Vladislav006

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Найдем первую производную
$(x+cos \frac{1}{x^2})' = (x)'+(cos \frac{1}{x^2})'=1-sin\frac{1}{x^2}*(-2x^{-3})=$
$= 1+(2x^{-3})*sin\frac{1}{x^2}$

Найдем вторую производную
$(1+(2x^{-3})*sin\frac{1}{x^2})' = (1)'+((2x^{-3})*sin\frac{1}{x^2})' = ((2x^{-3})*sin\frac{1}{x^2})'=$
$= (2x^{-3})'*sin\frac{1}{x^2} + (2x^{-3})* (sin\frac{1}{x^2})'=$
$= -6x^{-4} * sin\frac{1}{x^2} + 2x^{-3}* cos\frac{1}{x^2}*(-2x^{-3}) =$
$= -6x^{-4} * sin\frac{1}{x^2} - 4x^{-6}* cos\frac{1}{x^2}$

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hcosxkak reshit

ctgy22 najti yxpomogite pozhalujsta

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