Last updated at Dec. 16, 2024 by Teachoo
Ex 12.2, 10 Find the derivative of cos x from first principle. Let f (x) = cos x We need to find fβ(x) We know that fβ(x) = (πππ)β¬(ββ0) πβ‘γ(π₯ + β) β π(π₯)γ/β Here, f (x) = cos x So, f (x + h) = cos (x + h) Putting values, fβ (x) = limβ¬(hβ0)β‘γ(πππ (π + π) βγ πππγβ‘π)/hγ Using cos A β cos B = β 2 sin ((π΄ + π΅)/2) sin ((π΄ β π΅)/2) = limβ¬(hβ0)β‘γ(βπ πππ((π + (π + π))/π) . πππ(((π + π) β π)/π))/hγ = limβ¬(hβ0)β‘γ(β2 π ππ((2π₯ + β)/2) . π ππ(β/2))/hγ = limβ¬(hβ0)β‘γβ2 sinβ‘((2π₯ + β)/2).γsin γβ‘γβ/2γ/βγ = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2).γsin γβ‘γβ/2γ/(β/2)γ Using (πππ)β¬(π₯β0)β‘γ π ππβ‘π₯/π₯γ=1 Replacing x by β/2 β (πππ)β¬(ββ0) π ππβ‘γ β/2γ/(( β)/2) = 1 = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2).(π₯π’π¦)β¬(π‘βπ) γπ¬π’π§ γβ‘γπ/πγ/(π/π)γ = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2).πγ = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2) γ Putting h = 0 = βsinβ‘((2π₯ +0)/2) = βsinβ‘(2π₯/2) = β sin x β΄ fβ(x) = βsin x
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About the Author
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 and Computer Science at Teachoo