Question

Help !!!
See question in image.
Please show workings .
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Answer 1

Answer:

see explanation

Step-by-step explanation:

Given f(x) then the derivative f'(x) is

f'(x) = lim( h tends to 0 ) $\frac{f(x+h)-f(x)}{h}$

      = lim( h to 0 )$\frac{x+h+\frac{1}{x+h }-(x+\frac{1}{x} ) }{h}$

      = lim(h to 0) $\frac{x+h+\frac{1}{x+h}-x-\frac{1}{x} }{h}$

      = (lim(h to 0) $\frac{h+\frac{1}{x+h}-\frac{1}{x} }{h}$

      = lim( h to 0 ) $\frac{hx(x+h)+x-(x+h)}{hx(x+h)}$

      = lim( h to 0 ) $\frac{hx(x+h)+x-x-h}{hx(x+h)}$

      = lim(h to 0 ) $\frac{hx(x+h)}{hx(x+h)}$ - $\frac{h}{hx(x+h)}$

Cancel the numerator/denominator of first fraction and h in the second

= lim ( h to 0 ) 1 - $\frac{1}{x(x+h)}$ ( let h go to zero ), then

f'(x) = 1 - $\frac{1}{x^2}$