[tex]f(x) = \dfrac{5}{x} =5x^-^1[/tex]
[tex]f'(x)= -5x^-^2= \dfrac{-5}{x^2} [/tex]
[tex]\text{plug in x = 2 and x=4}[/tex]
[tex]f'(2)= \dfrac{-5}{2^2} = 2^2 = 4 = \dfrac{-5}{4} [/tex]
[tex]f'(4)= \dfrac{-5}{4^2} = 4^2 = 4*4 = 16 = \dfrac{-5}{16} [/tex]
[tex]\text{Hence solved }[/tex]
