Автор - NekoDuda

Помогите решить производные

1) y = sqrt[5]{x} +xsqrt[3]{x}

2) y= cos in(1-x^2)

3) y=sinsqrt{3} +1sin^23x/3cos6x

Ответ

Проверено экспертом

Автор - Universalka

1)y=sqrt[5]{x}+xsqrt[3]{x}=x^{frac{1}{5}}+x*x^{frac{1}{3}}=x^{frac{1}{5}}+x^{frac{4}{3}}\\y'=(x^{frac{1}{5} })'+(x^{frac{4}{3}})'=frac{1}{5}x^{-frac{4}{5}}+frac{4}{3}x^{frac{1}{3}}=frac{1}{5sqrt[5]{x^{4}}}+frac{4sqrt[3]{x}}{3}

2)y=Cos(1-x^{2} )\\y'=[Cos(1-x^{2})]'=-Sin(1-x^{2})*(1-x^{2})'=-Sin(1-x^{2})*(-2x)=2xSin(1-x^{2})

3)Sinsqrt{3}+frac{1}{3}*frac{Sin^{2}3x }{Cos6x}=(Sinsqrt{3})'+frac{1}{3}(frac{Sin^{2}3x }{Cos6x})'=Cossqrt{3} *(sqrt{3})'+frac{1}{3}frac{2Sin3x*(Sin3x)'*Cos6x-Sin^{2}3x*(Cos6x)' }{Cos^{2}6x}=frac{1}{3}frac{2Sin3x*Cos3x*(3x)'+Sin^{2}3xSin6x*(6x)' }{Cos^{2}6x }=frac{1}{3}*frac{3Sin6x+6Sin^{2}3x*Sin6x }{Cos^{2}6x } =frac{Sin6x(1+2Sin^{2}3x) }{Cos^{2}6x}

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