Question

срочно)))
алгебра 10 класс

Answer 1

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)

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

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)

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

з)

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