Derivatives by formula - sin & cos

Chapter 12 Class 11 Limits and Derivatives
Concept wise

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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) = (πππ)β¬(ββ0) πβ‘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¬(hβ0)β‘γ(πππ (π + π) βγ πππγβ‘π)/hγ Using cos A β cos B = β 2 sin ((π΄ + π΅)/2) sin ((π΄ β π΅)/2) = limβ¬(hβ0)β‘γ(βπ πππ((π + (π + π))/π) . πππ(((π + π) β π)/π))/hγ = limβ¬(hβ0)β‘γ(β2 π ππ((2π₯ + β)/2) . π ππ(β/2))/hγ = limβ¬(hβ0)β‘γβ2 sinβ‘((2π₯ + β)/2).γsin γβ‘γβ/2γ/βγ = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2).γsin γβ‘γβ/2γ/(β/2)γ Using (πππ)β¬(π₯β0)β‘γ π ππβ‘π₯/π₯γ=1 Replacing x by β/2 β (πππ)β¬(ββ0) π ππβ‘γ β/2γ/(( β)/2) = 1 = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2).(π₯π’π¦)β¬(π‘βπ) γπ¬π’π§ γβ‘γπ/πγ/(π/π)γ = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2).πγ = limβ¬(hβ0)β‘γβsinβ‘((2π₯ + β)/2) γ Putting h = 0 = βsinβ‘((2π₯ +0)/2) = βsinβ‘(2π₯/2) = β sin x β΄ fβ(x) = βsin x