$ 
				\eqalign{ 
				&f(x)=\frac{1} 
				{{x^2}}\cr 
				&f(x)=x^{-2}\cr 
				&f\,'(x)=-2\cdot x^{-3}\cr 
				&f\,'(x)=-\frac{2} 
				{{x^3}}\cr} 
				$ | 
			
				$ 
				\eqalign{ 
				&g(x)=\sqrt x\cr 
				&g(x)=x^{\frac{1} 
				{2}}\cr 
				&g\,'(x)=\frac{1} 
				{2}x^{-\frac{1} 
				{2}}\cr 
				&g\,'(x)=\frac{1} 
				{2}\cdot\frac{1} 
				{{x^{\frac{1} 
				{2}}}}\cr 
				&g\,'(x)=\frac{1} 
				{2}\cdot\frac{1} 
				{{\sqrt x}}\cr 
				&g\,'(x)=\frac{1} 
				{{2\sqrt x}}\cr} 
				$ | 
			
				$ 
				\eqalign{ 
				&h(x)=\frac{2} 
				{{\root3\of{x^2}}}\cr 
				&h(x)=\frac{2} 
				{{x^{\frac{2} 
				{3}}}}\cr 
				&h(x)=2\cdot x^{-\frac{2} 
				{3}}\cr 
				&h\,'(x)=2\cdot-\frac{2} 
				{3}x^{-1\frac{2} 
				{3}}\cr 
				&h\,'(x)=-\frac{4} 
				{3}x^{-\frac{5} 
				{3}}\cr 
				&h\,'(x)=-\frac{4} 
				{3}\cdot\frac{1} 
				{{x^{\frac{5} 
				{3}}}}\cr 
				&h\,'(x)=-\frac{4} 
				{3}\cdot\frac{1} 
				{{\root3\of{x^5}}}\cr 
				&h\,'(x)=-\frac{4} 
				{{3\root3\of{x^5}}}\cr} 
				$ | 
			
				 
					$ 
					\eqalign{ 
					&k(x)=\frac{{x^3-4}} 
					{{3x}}\cr 
					&k(x)=\frac{{x^3}} 
					{{3x}}-\frac{4} 
					{{3x}}\cr 
					&k(x)=\frac{1} 
					{3}x^2-\frac{4} 
					{3}x^{-1}\cr 
					&k\,'(x)=\frac{1} 
					{3}\cdot2x-\frac{4} 
					{3}\cdot-1\cdot x^{-2}\cr 
					&k\,'(x)=\frac{2} 
					{3}x+\frac{4} 
					{3}\cdot\frac{1} 
					{{x{}^2}}\cr 
					&k\,'(x)=\frac{2} 
					{3}x+\frac{4} 
					{{3x{}^2}}\cr}$ 
					$\eqalign{ 
					&k\,'(x)=\frac{2} 
					{3}x\cdot\frac{{x^2}} 
					{{x^2}}+\frac{4} 
					{{3x^2}}\cr 
					&k\,'(x)=\frac{{2x^3}} 
					{{3x^2}}+\frac{4} 
					{{3x^2}}\cr 
					&k\,'(x)=\frac{{2x^3+4}} 
					{{3x^2}}\cr} 
					$ 
			 |