Ответ:
[tex]\displaystyle y' =-\frac{20}{(x^2-10)^2}[/tex]
[tex]\displaystyle y'(3)=-20[/tex]
Пошаговое объяснение:
3) Найдите производную функции:
[tex]\displaystyle \bf y=\frac{x^2}{x^2-10}[/tex]
[tex]\displaystyle \bf \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}[/tex]
[tex]\displaystyle y'=\frac{(x^2)'(x^2-10)-x^2(x^2-10)'}{(x^2-10)^2} =\frac{2x(x^2-10)-x^2\cdot2x}{(x^2-10)^2} =\\\\=\frac{2x^3-20x-2x^3}{(x^2-10)^2} =-\frac{20}{(x^2-10)^2}[/tex]
Найдем y'(3):
[tex]\displaystyle y'(3)=-\frac{20}{(3^2-10)^2}=-20[/tex]
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Verified answer
Ответ:
[tex]\displaystyle y' =-\frac{20}{(x^2-10)^2}[/tex]
[tex]\displaystyle y'(3)=-20[/tex]
Пошаговое объяснение:
3) Найдите производную функции:
[tex]\displaystyle \bf y=\frac{x^2}{x^2-10}[/tex]
[tex]\displaystyle \bf \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}[/tex]
[tex]\displaystyle y'=\frac{(x^2)'(x^2-10)-x^2(x^2-10)'}{(x^2-10)^2} =\frac{2x(x^2-10)-x^2\cdot2x}{(x^2-10)^2} =\\\\=\frac{2x^3-20x-2x^3}{(x^2-10)^2} =-\frac{20}{(x^2-10)^2}[/tex]
Найдем y'(3):
[tex]\displaystyle y'(3)=-\frac{20}{(3^2-10)^2}=-20[/tex]