Question

Пожалуйста,срочно надо

Answer 1

$a)\;f(x)=2x^6-x^4+3x^3+18\\f'(x)=16x^5-4x^3+9x^2\\\\b)\;y=x^2(x^5+2x^2-7x)=x^7+2x^4-7x^3\\y'=7x^6+8x^3-21x^2\\\\c)\;f(x)=\frac6x\\f'(x)=-\frac6{x^2}\\f'(x)=-6\\\\d)\;f(x)=\frac{7-2x^2}{7+2x^2}\\f'(x)=\frac{-4x(7+2x^2)-4x(7-2x^2)}{(7+2x^2)^2}=\frac{-28x-8x^3-28x+8x^3}{(7+2x^2)^2}=-\frac{56x}{(7+2x^2)^2}\\f'(0)=0\\f'(1)=-\frac{56}{(7+2)^2}=-\frac{56}{81}$

$e)\;f(x)=(x^2-4x)(4x^2-x)=4x^4-17x^3+4x^2\\f'(x)=16x^3-51x^2+8x\\\\f)\;f(x)=\frac{4x^2-17x+4}{x-4}\\f'(x)=\frac{(4x-17)(x-4)-(4x^2-17x+4)}{(x-4)^2}=\frac{4x^2-33x+68-4x^2+17x-4}{(x-4)^2}=\frac{64-16x}{(x-4)^2}=\\=\frac{-16(x-4)}{(x-4)^2}=-\frac{16}{x-4}$