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Ex 13.2
Ex 13.2, 2
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Ex 13.2, 4 (i) Important
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Ex 13.2, 4 (iii) Important
Ex 13.2, 4 (iv)
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Ex 13.2, 7 (i) Important
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Ex 13.2, 9 (i)
Ex 13.2, 9 (ii) Important
Ex 13.2, 9 (iii)
Ex 13.2, 9 (iv) Important
Ex 13.2, 9 (v)
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Ex 13.2, 10 Important You are here
Ex 13.2, 11 (i)
Ex 13.2, 11 (ii) Important
Ex 13.2, 11 (iii) Important
Ex 13.2, 11 (iv)
Ex 13.2, 11 (v) Important
Ex 13.2, 11 (vi)
Ex 13.2, 11 (vii) Important
Last updated at March 30, 2023 by Teachoo
Ex 13.2, 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