Miscellaneous

Chapter 12 Class 11 Limits and Derivatives
Serial order wise

### Transcript

Misc 1 Find the derivative of the following functions from first principle: (ii) (βπ₯)^(β1) Let f (x) = (βπ₯)^(β1) We need to find Derivative of f(x) i.e. fβ (x) We know that fβ(x) = limβ¬(hβ0) fβ‘γ(x + h) β f(x)γ/h Here, f (x) = (βπ₯)^(β1) So, f (x + h) = (β(π₯+β))^(β1) Putting values fβ(x) = limβ¬(hβ0)β‘γ(γ(β(π₯ + β)) γ^(β1) γβ (β π₯)γ^(β1))/βγ = limβ¬(hβ0)β‘γ((1/(β(π₯ + β))) β(1/(βπ₯)))/βγ = limβ¬(hβ0)β‘γ((β1)/(π₯ + β) + 1/π₯)/βγ = limβ¬(hβ0)β‘γ((β π₯ + π₯ + β)/(π₯ (π₯ + β) ))/βγ = limβ¬(hβ0)β‘γ(β π₯ + π₯ + β)/(βπ₯ (π₯ + β))γ = limβ¬(hβ0)β‘γβ/(βπ₯ (π₯ + β))γ = limβ¬(hβ0)β‘γ1/(π₯ (π₯ + β))γ Putting h = 0 = 1/(π₯ (π₯ + 0)) = π/ππ

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.