Question

Нужно найти производную функции 1)u(x)=(x-5)*(2x-5) 2)f(x)=(2x+1)/(2x-1) 3)y(x)=x+√x 4)y(x)=(5+√x)*(√x-5)+√x 5)f(x)=(3x+5)/(5x+3) 6)y(x)=(1+x)(x-1)+6x^4

Answer 1

u'=(2x-5)+2(x-5)=4x-15

F'=1/(2x-1)^2*[2(2x-1)-2(2x+1)]=-4/(2x-1)^2

y'=1+1/2sqrt(x)

y=(x-25)+sqrt(x) y'=1+1/2sqrt(x)

f'=(3(5x+3)-(3x+5)*5)/(5x+3)^2=-16/(5x+3)^2

y=(x^2-1)+6x^4  y'=2x-1+24x^3

Answer 2

1.

$\\u(x)=(x-5)(2x-5)\\ u'(x)=2x-5+(x-5)\cdot2\\ u'(x)=2x-5+2x-10\\ u'(x)=4x-15$

2.

$\\f(x)=\frac{2x+1}{2x-1}\\ f'(x)=\frac{2(2x-1)-(2x+1)\cdot2}{(2x-1)^2}\\ f'(x)=\frac{4x-2-4x-2}{(2x-1)^2}\\ f'(x)=-\frac{4}{(2x-1)^2}$

3.

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

4.

$\\y(x)=(5+\sqrt x)(\sqrt x-5)+\sqrt x\\ y(x)=x-25+\sqrt x\\ y'(x)=1+\frac{1}{2\sqrt x}\\$

5.

$\\f(x)=\frac{3x+5}{5x+3}\\ f'(x)=\frac{3(5x+3)-(3x+5)\cdot 5}{(5x+3)^2}\\ f'(x)=\frac{15x+9-15x-25}{(5x+3)^2}\\ f'(x)=-\frac{16}{(5x+3)^2}$

6.

$\\y(x)=(1+x)(x-1)+6x^4\\ y(x)=x^2-1+6x^4\\ y'(x)=2x+24x^3\\$