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Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 10

Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 11
Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 12

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Ex 13.2, 4 Find the derivative of the following functions from first principle. (iv) (π‘₯ + 1)/(π‘₯ βˆ’ 1) Let f (x) = (π‘₯ + 1)/(π‘₯ βˆ’ 1) We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = lim┬(hβ†’0) f⁑〖(x + h) βˆ’ f(x)γ€—/h Here, f (x) = (π‘₯ + 1)/(π‘₯ βˆ’ 1) So, f (x + h) = ((π‘₯ + β„Ž) + 1)/((π‘₯ + β„Ž) βˆ’ 1) Ex 13.2, 4 Find the derivative of the following functions from first principle. (iv) (π‘₯ + 1)/(π‘₯ βˆ’ 1) Let f (x) = (π‘₯ + 1)/(π‘₯ βˆ’ 1) We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = lim┬(hβ†’0) f⁑〖(x + h) βˆ’ f(x)γ€—/h Here, f (x) = (π‘₯ + 1)/(π‘₯ βˆ’ 1) So, f (x + h) = ((π‘₯ + β„Ž) + 1)/((π‘₯ + β„Ž) βˆ’ 1) Putting values f’(x) = lim┬(hβ†’0)⁑〖([((π‘₯ + β„Ž) + 1)/(π‘₯ + β„Ž βˆ’ 1)] βˆ’[ (π‘₯ + 1)/(π‘₯ βˆ’ 1)])/hγ€— = lim┬(hβ†’0)⁑〖((π‘₯ + β„Ž + 1)/(π‘₯ + β„Ž βˆ’ 1) βˆ’ (π‘₯ + 1)/(π‘₯ βˆ’ 1))/hγ€— = lim┬(hβ†’0)⁑〖(((π‘₯ βˆ’ 1)(π‘₯ + β„Ž + 1) βˆ’ (π‘₯ + 1)( π‘₯ + β„Ž βˆ’ 1))/(( π‘₯ + β„Ž βˆ’ 1 ) (π‘₯ βˆ’ 1)))/β„Žγ€— = lim┬(hβ†’0)⁑〖((π‘₯ βˆ’ 1) ((π‘₯ + 1) + β„Ž) βˆ’ (π‘₯ + 1)( (π‘₯ βˆ’ 1) + β„Ž))/(β„Ž( π‘₯ + β„Ž βˆ’ 1 ) (π‘₯ βˆ’ 1))γ€— = lim┬(hβ†’0)⁑〖((π‘₯ βˆ’ 1)(π‘₯ + 1) + (π‘₯ βˆ’1)β„Ž βˆ’ (π‘₯ + 1)(π‘₯ βˆ’ 1) βˆ’ (π‘₯ + 1) β„Ž)/(β„Ž( π‘₯ + β„Ž βˆ’ 1 ) (π‘₯ βˆ’ 1))γ€— = lim┬(hβ†’0)⁑〖((π‘₯2 βˆ’ 1) + π‘₯β„Ž βˆ’ (π‘₯2 βˆ’ 1) βˆ’ π‘₯β„Ž βˆ’ β„Ž)/(β„Ž( π‘₯ + β„Ž βˆ’ 1 ) (π‘₯ βˆ’ 1))γ€— = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖(βˆ’ 2β„Ž )/(β„Ž (π‘₯ + β„Ž βˆ’ 1) (π‘₯ βˆ’ 1))γ€— = lim┬(hβ†’0)⁑〖(βˆ’2)/((π‘₯ + β„Ž βˆ’ 1) (π‘₯ βˆ’ 1))γ€— Putting h = 0 = (βˆ’2)/((π‘₯ + 0 βˆ’ 1)(π‘₯ βˆ’ 1)) = (βˆ’2)/((π‘₯ βˆ’ 1) (π‘₯ βˆ’ 1)) = (βˆ’2)/(π‘₯ βˆ’ 1)^2 Hence, f’(x) = (βˆ’πŸ)/(𝒙 βˆ’ 𝟏)^𝟐

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