The definition says[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]###With [tex]f(x)=(x+2)^2+8[/tex], we get[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{((x+h+2)^2+8)-((x+2)^2+8)}h=\lim_{h\to0}\frac{(x+2)^2+2h(x+2)+h^2+8-(x+2)^2-8}h[/tex][tex]f'(x)=\displaystyle\lim_{h\to0}\frac{2h(x+2)+h^2}h=\lim_{h\to0}2(x+2)+h=2(x+2)[/tex]###With [tex]f(x)=\sqrt{2x-5}[/tex]:[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\sqrt{2(x+h)-5}-\sqrt{2x-5}}h=\lim_{h\to0}\frac{(2(x+h)-5)-(2x-5)}{h(\sqrt{2(x+h)-5}+\sqrt{2x-5})}[/tex][tex]f'(x)=\displaystyle\lim_{h\to0}\frac{2h}{h(\sqrt{2x+2h-5}+\sqrt{2x-5})}=\lim_{h\to0}\frac2{\sqrt{2x+2h-5}+\sqrt{2x-5}}[/tex][tex]f'(x)=\dfrac2{\sqrt{2x-5}+\sqrt{2x-5}}=\dfrac1{\sqrt{2x-5}}[/tex]###With [tex]f(x)=\dfrac1{3x-1}[/tex]:[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{3(x+h)-1}-\frac1{3x-1}}h=\lim_{h\to0}\frac{(3x-1)-(3(x+h)-1)}{h(3(x+h)-1)(3x-1)}[/tex][tex]f'(x)=\displaystyle\lim_{h\to0}\frac{-3h}{h(3x+3h-1)(3x-1)}=\lim_{h\to0}\frac{-3}{(3x+3h-1)(3x-1)}[/tex][tex]f'(x)=-\dfrac3{(3x-1)^2}[/tex]