$a){\left(2x\sqrt{x}\right)}^{{}^{\prime }}={\left(2x\right)}^{{}^{\prime }}\cdot \sqrt{x}+2x\cdot {\left(\sqrt{x}\right)}^{{}^{\prime }}=2\cdot \sqrt{x}+2x\cdot \frac{1}{2\sqrt{x}}=2\sqrt{x}+\frac{x}{\sqrt{x}}=2\sqrt{x}+\sqrt{x}=3\sqrt{x}$  

 

$b){\left(-4x^6\sqrt{x}\right)}^{{}^{\prime }}={\left(-4x^6\right)}^{{}^{\prime }}\cdot \sqrt{x}-4x^6\cdot {\left(\sqrt{x}\right)}^{{}^{\prime }}=-4\cdot 6x^5\cdot \sqrt{x}-4x^6\cdot \frac{1}{2\sqrt{x}}=-24x^5\sqrt{x}-\frac{2x^6}{\sqrt{x}}=-24x^{\frac{11}{2}}-2x^{\frac{11}{2}}=-26x^{\frac{11}{2}}$  

 

$c){\left[\left(2x^3-4\right)\left(x^4+x\right)\right]}^{{}^{\prime }}=\left(2x^3-4\right){}^{\prime }\cdot \left(x^4+x\right)+\left(2x^3-4\right)\cdot {\left(x^4+x\right)}^{{}^{\prime }}=\left({\left(2x^3\right)}^{{}^{\prime }}-{\left(4\right)}^{{}^{\prime }}\right)\cdot \left(x^4+x\right)+\left(2x^3-4\right)\left({\left(x^4\right)}^{{}^{\prime }}+{\left(x\right)}^{{}^{\prime }}\right)=$ $c){\left[\left(2x^3-4\right)\left(x^4+x\right)\right]}^{{}^{\prime }}=\left(2x^3-4\right){}^{\prime }\cdot \left(x^4+x\right)+\left(2x^3-4\right)\cdot {\left(x^4+x\right)}^{{}^{\prime }}=\left({\left(2x^3\right)}^{{}^{\prime }}-{\left(4\right)}^{{}^{\prime }}\right)\cdot \left(x^4+x\right)+\left(2x^3-4\right)\left({\left(x^4\right)}^{{}^{\prime }}+{\left(x\right)}^{{}^{\prime }}\right)=$  

$=6x^2\cdot \left(x^4+x\right)+\left(2x^3-4\right)\left(4x^3+1\right)=6x^6+6x^3+8x^6+2x^3-16x^3-4=14x^6-8x^3-4$  

 

$d){\left[\left(x^3+x^2-4\right)\left(4x^4-x^2\right)\right]}^{{}^{\prime }}={\left(x^3+x^2-4\right)}^{{}^{\prime }}\left(4x^4-x^2\right)+\left(x^3+x^2-4\right){\left(4x^4-x^2\right)}^{{}^{\prime }}=\left(3x^2+2x\right)\left(4x^4-x^2\right)+\left(x^3+x^2-4\right)\left(16x^3-2x\right)=$  

$=x\left(3x+2\right)\cdot x^2\left(4x^2-1\right)+\left(x^3+x^2-4\right)\cdot 2x\left(8x^2-1\right)=x^3\left(12x^3-3x+8x^2-2\right)+2x\left(8x^5-x^3+8x^4-x^2-32x^2+4\right)=$  

$=12x^6-3x^4+8x^5-2x^3+2x\left(8x^5+8x^4-x^3-33x^2+4\right)=12x^6+8x^5-3x^4-2x^3+16x^6+16x^5-2x^4-66x^3+8x=$  

$=28x^6+24x^5-5x^4-68x^3+8x=x\left(28x^5+24x^4-5x^3-68x^2+8\right)$  

 

$e){\left[\sqrt{x}\left(3x^5-x^2\right)\right]}^{{}^{\prime }}={\left(\sqrt{x}\right)}^{{}^{\prime }}\left(3x^5-x^2\right)+\sqrt{x}{\left(3x^5-x^2\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{x}}\left(3x^5-x^2\right)+\sqrt{x}\left(15x^4-2x\right)=\frac{3}{2}{x}^{\frac{9}{2}}-\frac{1}{2}{x}^{\frac{3}{2}}+15x^{\frac{9}{2}}-2x^{\frac{3}{2}}=16\frac{1}{2}{x}^{\frac{9}{2}}-2\frac{1}{2}{x}^{\frac{3}{2}}=\frac{1}{2}{x}^{\frac{3}{2}}\left(33x^3-5\right)$  

 

$f){\left[\sqrt{x}\left(x^4+4\sqrt{x}\right)\right]}^{{}^{\prime }}={\left(\sqrt{x}\right)}^{{}^{\prime }}\left(x^4+4\sqrt{x}\right)+\sqrt{x}{\left(x^4+4\sqrt{x}\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{x}}\left(x^4+4\sqrt{x}\right)+\sqrt{x}\left(4x^3+4\cdot \frac{1}{2\sqrt{x}}\right)=$  

$=\frac{1}{2}{x}^{\frac{7}{2}}+2+4x^{\frac{7}{2}}+2=4\frac{1}{2}{x}^{\frac{7}{2}}+4$