Example 19 (ii) - Chapter 12 Class 11 Limits and Derivatives

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Transcript

Example 19 Find the derivative of f from the first principle, where f is given by (ii) f(x) = x + 1/x Given f (x) = x + 1/x We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = lim┬(hβ†’0) 𝑓⁑〖(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)γ€—/β„Ž Here, f(x) = x + 1/x f(x + h) = (x + h) + 1/(x + β„Ž) Putting values f’(x) = lim┬(hβ†’0)⁑〖(((π‘₯ + β„Ž)+ 1/(π‘₯ + β„Ž)) βˆ’ (π‘₯ + 1/π‘₯))/hγ€— = lim┬(hβ†’0)⁑〖(π‘₯ + β„Ž+ 1/(π‘₯ + β„Ž) βˆ’ π‘₯ βˆ’ 1/(π‘₯ ))/hγ€— = lim┬(hβ†’0)⁑〖(β„Ž+ 1/(π‘₯ + β„Ž) βˆ’ 1/π‘₯ + π‘₯ βˆ’ π‘₯)/β„Žγ€— = lim┬(hβ†’0)⁑〖(β„Ž+ 1/(π‘₯ + β„Ž) βˆ’ 1/π‘₯)/β„Žγ€— = lim┬(hβ†’0)⁑〖(β„Ž+ (π‘₯ βˆ’ (π‘₯ βˆ’ β„Ž))/((π‘₯ + β„Ž) (π‘₯)))/β„Žγ€— = lim┬(hβ†’0)⁑〖(β„Ž + ((βˆ’β„Ž))/((π‘₯ + β„Ž) π‘₯))/β„Žγ€— = lim┬(hβ†’0)⁑〖( β„Ž(1 βˆ’ 1/((π‘₯ + β„Ž) π‘₯)))/β„Žγ€— = lim┬(hβ†’0)⁑(1βˆ’1/((π‘₯ + β„Ž) π‘₯)) Putting h = 0 = [1βˆ’1/(π‘₯ (π‘₯ + 0) )] = πŸβˆ’πŸ/(π’™πŸ )

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