Question

Найдите производную функции: f(x)=(8x^5-5x^8)^12 f(x)=(4x^10-5x)^10 f(x)=(1/9-3x^3)^27 f(x)=(x^5-4x^4)^130

Answer 1

$\\f(x)=(8x^5-5x^8)^{12}\\ f'(x)=12(8x^5-5x^8)^{11}(40x^4-40x^7)\\ f'(x)=480x^{59}(8-5x^3)^{11}(1-x^3)\\ f'(x)=480x^{59}(5x^3-8)^{11}(x^3-1)$

 

$\\f(x)=(4x^{10}-5x)^{10}\\ f'(x)=10(4x^{10}-5x)^9(40x^9-5)\\ f'(x)=50x^9(4x^9-5)^9(8x^9-1)$

 

$\\f(x)=(\frac{1}{9}-3x^3)^{27}\\ f'(x)=27(\frac{1}{9}-3x^3)^{26}\cdot(-9x^2)\\ f'(x)=-243x^2(3x^3-\frac{1}{9})^{26}$

 

$\\f(x)=(x^5-4x^4)^{130} \\ f'(x)=130(x^5-4x^4)^{129}(5x^4-16x^3)\\ f'(x)=130x^{519}(x-4)^{129}(5x-16)$

Answer 2

1) 12*(8x^5-5x^8)^11*(40x^4-40x^7)=480*x^(55)(8-5x^3)^11*x^4(1-x^3)=

480*x^(59)(8-5x^3)(^11)(1-x^3)

2) 10*(4x^10-5x)^9*(40x^9-5)=50*x^9(4x^9-5)^(9)*(8x^9-1)

3) 27(1/9-3x^3)^26*(-9x^2)=-x^2*(1/9-3x^3)^26*243

4)(x^5-4x^4)^129*(5x^4-4x^3)