Examples

Chapter 12 Class 11 Limits and Derivatives
Serial order wise

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. fβ (x) We know that fβ(x) = limβ¬(hβ0) πβ‘γ(π₯ + β) β π(π₯)γ/β Here, f(x) = x + 1/x f(x + h) = (x + h) + 1/(x + β) Putting values fβ(x) = limβ¬(hβ0)β‘γ(((π₯ + β)+ 1/(π₯ + β)) β (π₯ + 1/π₯))/hγ = limβ¬(hβ0)β‘γ(π₯ + β+ 1/(π₯ + β) β π₯ β 1/(π₯ ))/hγ = limβ¬(hβ0)β‘γ(β+ 1/(π₯ + β) β 1/π₯ + π₯ β π₯)/βγ = limβ¬(hβ0)β‘γ(β+ 1/(π₯ + β) β 1/π₯)/βγ = limβ¬(hβ0)β‘γ(β+ (π₯ β (π₯ β β))/((π₯ + β) (π₯)))/βγ = limβ¬(hβ0)β‘γ(β + ((ββ))/((π₯ + β) π₯))/βγ = limβ¬(hβ0)β‘γ( β(1 β 1/((π₯ + β) π₯)))/βγ = limβ¬(hβ0)β‘(1β1/((π₯ + β) π₯)) Putting h = 0 = [1β1/(π₯ (π₯ + 0) )] = πβπ/(ππ )