$f'\left(x\right)=\frac{1}{x}$  

 $f'\left(x\right)=-\frac{1}{\text{}x^2}$  

 $f'\left(x\right)=\frac{1}{\text{x}^2}$  

 $f'\left(x\right)=-\frac{1}{x}$  

 $f'\left(x\right)=2\sqrt{x}$  

 $f'\left(x\right)=\frac{1}{\sqrt{x}}$  

 $f'\left(x\right)=\frac{1}{2\sqrt{x}}$  

 $f'\left(x\right)=-\frac{1}{\sqrt{x}}$  

 $f'\left(x\right)=2019x^{2020}$  

 $f'\left(x\right)=2020x^{2019}$  

 $f'\left(x\right)=2020$  

 $f'\left(x\right)=2019$  

 $f'\left(x\right)=\frac{5}{6}x^{\frac{1}{6}}$  

 $f'\left(x\right)=\frac{6}{5}x^{\frac{1}{6}}$  

 $f'\left(x\right)=\frac{5}{6x^{\frac{1}{6}}}$  

 $f'\left(x\right)=\frac{6}{5x^{\frac{1}{6}}}$  

 $f\left(x\right)=-\frac{21}{x^6}$  

 $f\left(x\right)=-\frac{21}{x^8}$  

 $f\left(x\right)=\frac{21}{x^6}$  

 $f\left(x\right)=\frac{21}{x^8}$  

$\phi$

10

0

10x

 $6x^2+7$  

 $6x^2+12x-5$  

 $7$  

 $6x-7$  

 $\frac{\left(3x+1\right)\left(10x\right)-\left(5x^2+2\right)\left(3\right)}{\left(3x+1\right)^2}$  

 $\frac{\left(3x+1\right)\left(5x\right)-\ \left(5x\ +2\right)}{\left(3x+1\right)^2}$  

 $\frac{\left(5x+1\right)^2\left(2\right)-\ 6x\left(2\right)}{\left(5x+2\right)^2}$  

 $\frac{\left(5x+2\right)}{\left(3x+1\right)^2}$  

 $\left(8x\ +2\ -\ 12x^2\right)\left(5x^2\ +\ -1\right)$  

 $\left(8x^2+3\right)\left(x^3+x\ +x^2\right)$  

 $\left(x^2+x^3+x\right)\left(16x\right)+\left(8x^2+3\right)\left(2x\ +3x^2\ +1\right)$  

 $\frac{\left(5x\ -\ 1\right)}{2x}$  

-2

0

1

2