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Amerigokz
@Amerigokz
August 2022
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100 баллов
Найти частные производные первого и второго порядка
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sedinalana
Verified answer
Z`(x)=3x²/cos²(x³+y²)
z`(y)=2y/cos²(x³+y²)
z``(xx)=[6x*cos²(x³+y²)+3x²*2cos(x³+y²)*sin(x³+y²)*3x²]/cos^4(x³+y²)=
=[6x*cos²(x³+y²)+9x^4sin(2x³+2y²)]/cos^4(x³+y²)
z``(xy)=[3x²*2cos(x³+y²)*sin(x²)*2y]/cos^4(x³+y²)=
=6x²y*sin(2x³+2y²)/cos^4(x³+y²)
z``(yx)=[2y*2cos(x³+y²)*sin(x³+y²)*2y]/cos^4(x³+y²)=
=2y²*sin(2x³+2y²)/cos^4(x³+y²)
z``(yy)=[2cos²(x³+y²)+2y*2cos(x³+y²)*sin(x³+y²)*2y]/cos^4(x³+y²)=
=[2cos²(x³+y²)+4y²*sin(2x³+2y²)]/cos^4(x³+y²)
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Answers & Comments
Verified answer
Z`(x)=3x²/cos²(x³+y²)z`(y)=2y/cos²(x³+y²)
z``(xx)=[6x*cos²(x³+y²)+3x²*2cos(x³+y²)*sin(x³+y²)*3x²]/cos^4(x³+y²)=
=[6x*cos²(x³+y²)+9x^4sin(2x³+2y²)]/cos^4(x³+y²)
z``(xy)=[3x²*2cos(x³+y²)*sin(x²)*2y]/cos^4(x³+y²)=
=6x²y*sin(2x³+2y²)/cos^4(x³+y²)
z``(yx)=[2y*2cos(x³+y²)*sin(x³+y²)*2y]/cos^4(x³+y²)=
=2y²*sin(2x³+2y²)/cos^4(x³+y²)
z``(yy)=[2cos²(x³+y²)+2y*2cos(x³+y²)*sin(x³+y²)*2y]/cos^4(x³+y²)=
=[2cos²(x³+y²)+4y²*sin(2x³+2y²)]/cos^4(x³+y²)