Найдите производную функции:[tex]y=sin^2*cos \sqrt{x} [/tex]. б)[tex]y= \frac{ \sqrt{ctgx} }{x^3} [/tex]. Created by alinaalina68. algebra-ru

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alinaalina68 @alinaalina68

July 2022 1 6 Report

Найдите производную функции:[tex]y=sin^2*cos \sqrt{x} [/tex]. б)[tex]y= \frac{ \sqrt{ctgx} }{x^3} [/tex]

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kmike21

Verified answer

A) $y'=2sinx*cosx*cos \sqrt{x} -sin^2x*sin \sqrt{x} * \frac{1}{2 \sqrt{x} }=$
= $sin2x*cos \sqrt{x} - \frac{sin^2x*sin \sqrt{x}}{2 \sqrt{x} }$
б) y'= $(x^{-3} *ctg^{ \frac{1}{2} }x)'=-3x^{-4}*ctg^{ \frac{1}{2} }x-x^{-3}* \frac{1}{2}* ctg^{ -\frac{1}{2} }x*sin^{-2}x=$
= $- \frac{3 \sqrt{ctgx} }{x^4} - \frac{1}{x^3* \sqrt{ctgx}*sin^2x }$

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Verified answer

22 ballasumma kornej uravneniya 2cosxcos2xcosx2cos2x1 prinadlezhashih promezh

tex s obyasneniemc6d89b8ce10b0ccfa49370d3ebcbf03c 39524

pomogite razgadat rebus0ff6b24e0b81d1c80b1ed749b0b3128a 71003

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