срочно))) алгебра 10 класс. Created by полинакласс. algebra-ru

a) y = x⁸y' = (x⁸)' = 8x⁷б) y = 7y' = 7' = 0в) y = 5x + 2y' = 5(x)' + 2' = 5 * 1 + 0 = 5г) д) y = 3Cosxy' = 3(Cosx)' = 3 * (- Sinx) = - 3Sinxе)ж)з)

срочно))) алгебра 10 класс...

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срочно)))
алгебра 10 класс

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Verified answer

a) y = x⁸

y' = (x⁸)' = 8x⁷

б) y = 7

y' = 7' = 0

в) y = 5x + 2

y' = 5(x)' + 2' = 5 * 1 + 0 = 5

г)

$y=4\sqrt{x}\\\\y'=4(\sqrt{x})'=4*\frac{1}{2\sqrt{x}}=\frac{2}{\sqrt{x} }$

д) y = 3Cosx

y' = 3(Cosx)' = 3 * (- Sinx) = - 3Sinx

е)

$y=7x^{5}-3x^{2}+\frac{5}{3x}+1=7x^{5}-3x^{2}+\frac{5}{3}x^{-1}+1\\\\y'=7(x^{5})'-3(x^{2})'+\frac{5}{3} (x^{-1})'+1'=7*5x^{4}-3*2x+\frac{5}{3}*(-x^{-2})+0=35x^{4}-6x-\frac{5}{3x^{2} }$

ж)

$y=\sqrt{x}(3x-1)\\\\y'=(\sqrt{x})'*(3x-1)+\sqrt{x}*(3x-1)'=\frac{1}{2\sqrt{x} }*(3x-1)+\sqrt{x}*3=\frac{3x-1}{2\sqrt{x} }+3\sqrt{x}=\frac{3x-1+6x}{2\sqrt{x} }=\frac{9x-1}{2\sqrt{x} }$

з)

$y=\frac{x^{2} }{x^{2}-1 }\\\\y'=\frac{(x^{2})'*(x^{2}-1 )-x^{2}*(x^{2} -1)'}{(x^{2}-1)^{2}}=\frac{2x*(x^{2}-1)-2x*x^{2}}{(x^{2}-1)^{2}}=\frac{2x^{3}-2x-2x^{3}}{(x^{2}-1)^{2}}=-\frac{2x}{(x^{2}-1)^{2}}$

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