Question

Помогите решить,пожалуйста

Answer 1

1) f'(x) = 91x^6 + 40x^4 - 15x^2 - 3
2) y(x) = 15x^19 - 10x^16
y'(x) = 295x^18 - 160x^5
3) f(x) = x² + 4/x
f'(x) = 2x - 1/x²

Answer 2

1
$f`(x)=91x^6-40x^4-15x^2-3$
2
$f`(x)=55x ^{10} *(3x^8-2x^5)+5x ^{11} *(24x^7-10x^4)=$$165 x^{18} -110 x^{15} +120 x^{18} -50 x^{15} =285 x^{18} -160 x^{15}$
3
$f`(x)=[(15x^4+12)*3x^2-6x(3x^5+12x)]/9x^4=$$(45x^6+36x^2-18x^6-72x^2)/9x^4=(27x^6-36x^2)/9x^4=$$9x^2*(3x^4-4)/9x^4=(3x^4-4)/x^2$