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

 

$b)f^{{}^{\prime }}\left(x\right)={\left({\left(1+x^2\right)}^4\right)}^{{}^{\prime }}=4{\left(1+x^2\right)}^3\cdot {\left(1+x^2\right)}^{{}^{\prime }}=4{\left(1+x^2\right)}^3\cdot 2x=8x{\left(1+x^2\right)}^3$  

 

$c)f^{{}^{\prime }}\left(x\right)={\left({\left(1-x\right)}^5\right)}^{{}^{\prime }}=5{\left(1-x\right)}^4\cdot {\left(1-x\right)}^{{}^{\prime }}=5{\left(1-x\right)}^4\cdot \left(-1\right)=-5{\left(1-x\right)}^4$ $c)f^{{}^{\prime }}\left(x\right)={\left({\left(1-x\right)}^5\right)}^{{}^{\prime }}=5{\left(1-x\right)}^4\cdot {\left(1-x\right)}^{{}^{\prime }}=5{\left(1-x\right)}^4\cdot \left(-1\right)=-5{\left(1-x\right)}^4$  

 

$d)f^{{}^{\prime }}\left(x\right)={\left({\left(4x^2+2\right)}^3\right)}^{{}^{\prime }}=3{\left(4x^2+2\right)}^2\cdot {\left(4x^2+2\right)}^{{}^{\prime }}=3{\left(4x^2+2\right)}^2\cdot 8x=24x{\left(4x^2+2\right)}^2$  

 

$e)f^{{}^{\prime }}\left(x\right)={\left({\left(3x+2\right)}^5\right)}^{{}^{\prime }}=5{\left(3x+2\right)}^4\cdot {\left(3x+2\right)}^{{}^{\prime }}=5{\left(3x+2\right)}^4\cdot 3=15{\left(3x+2\right)}^4$  

 

$f)f^{{}^{\prime }}\left(x\right)={\left({\left(5x^4-6x+2\right)}^7\right)}^{{}^{\prime }}=7{\left(5x^4-6x+2\right)}^6\cdot {\left(5x^4-6x+2\right)}^{{}^{\prime }}=7{\left(5x^4-6x+2\right)}^6\cdot \left(20x^3-6\right)$  

 

$g)f^{{}^{\prime }}\left(x\right)={\left(\sqrt{2x-4}\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{2x-4}}\cdot {\left(2x-4\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{2x-4}}\cdot 2=\frac{2}{2\sqrt{2x-4}}=\frac{1}{\sqrt{2x-4}}$  

 

$h)f^{{}^{\prime }}\left(x\right)={\left(\sqrt{6x^2+1}\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{6x^2+1}}\cdot {\left(6x^2+1\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{6x^2+1}}\cdot 12x=\frac{12x}{2\sqrt{6x^2+1}}=\frac{6x}{\sqrt{6x^2+1}}$  

 

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

 

$j)f^{{}^{\prime }}\left(x\right)={\left(\sqrt{5x^2-6x}\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{5x^2-6x}}\cdot {\left(5x^2-6x\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{5x^2-6x}}\cdot \left(10x-6\right)=\frac{10x-6}{2\sqrt{5x^2-6x}}=\frac{5x-3}{\sqrt{5x^2-6x}}$  

 

$k)f^{{}^{\prime }}\left(x\right)={\left(\sqrt{7x^2+x}\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{7x^2+x}}\cdot {\left(7x^2+x\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{7x^2+x}}\cdot \left(14x+1\right)=\frac{14x+1}{2\sqrt{7x^2+x}}$  

 

$l)f^{{}^{\prime }}\left(x\right)={\left(\sqrt{4-x^2}\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{4-x^2}}\cdot {\left(4-x^2\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{4-x^2}}\cdot \left(-2x\right)=-\frac{2x}{2\sqrt{4-x^2}}=-\frac{x}{\sqrt{4-x^2}}$