

Chapter 13 Class 11 Limits and Derivatives
Chapter 13 Class 11 Limits and Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Example 20 Find the derivative of f(x) from the first principle, where f(x) is (i) sin x + cos x Given f (x) = sin x + cos x We need to find Derivative of f(x) We know that fβ(x) = limβ¬(hβ0) πβ‘γ(π₯ + β) β π(π₯)γ/β Here, f (x) = sin x + cos x f (x + h) = sin (x + h) + cos (x + h) Putting values fβ(x) = limβ¬(hβ0)β‘γ(sinβ‘γ(π₯ + β)γ + cosβ‘(π₯ + β) β (sinβ‘π₯ + cosβ‘γπ₯)γ)/βγ Using sin (A + B) = sin A cos B + cos A sin B & cos (A + B) = cos A cos B β sin A sin B = limβ¬(hβ0)β‘γsinβ‘γπ₯ cosβ‘γβ +γ cosγβ‘γπ₯ sinβ‘γβ + cosβ‘γπ₯ cosβ‘γβ β sinβ‘γπ₯ γ sinγβ‘γβ βγ sinγβ‘γπ₯ βγ cosγβ‘π₯ γ γ γ γ γ γ γ γ γ/hγ = limβ¬(hβ0)β‘γcosβ‘γπ₯ sinβ‘γβ βγ sinγβ‘γπ₯ sinβ‘γβ + sinβ‘γπ₯ cosβ‘γβ β sinβ‘γπ₯ +γ cosγβ‘γπ₯ cosβ‘γβ βγ cosγβ‘π₯ γ γ γ γ γ γ γ γ γ/hγ = limβ¬(hβ0)β‘γsinβ‘γβ γ(cosγβ‘γπ₯ β sinβ‘γπ₯) + sinβ‘γπ₯ (cosβ‘γβ β 1) + cosβ‘π₯ (cosβ‘γβ β 1)γ γ γ γ γ γ/hγ = limβ¬(hβ0)β‘(sinβ‘γβ γ(cosγβ‘γπ₯ β sinβ‘γπ₯)γ γ γ/h+sinβ‘γπ₯ (cosβ‘γβ β 1)γ γ/h+cosβ‘γπ₯ (cosβ‘γβ β 1)γ γ/β)" " = limβ¬(hβ0)β‘γsinβ‘γβ γ(cosγβ‘γπ₯ βγ sinγβ‘γπ₯)γ γ γ/h+limβ¬(hβ0) sinβ‘γπ₯ (cosβ‘γβ β 1)γ γ/h+limβ¬(hβ0) cosβ‘γπ₯ (cosβ‘γβ β 1)γ γ/βγ = limβ¬(hβ0)β‘γ"(cos x β sin x)" sinβ‘β/β+limβ¬(hβ0) "(β sin x)" ((1 β cosβ‘γβ)γ)/βγ+limβ¬(hβ0) "(β cos x)" ((1 β cosβ‘γβ)γ)/β = "(cos x β sin x)" (π₯π’π¦)β¬(π‘βπ)β‘γπ¬π’π§β‘π/πβsinβ‘γπ₯ (π₯π’π¦)β¬(π‘βπ) γ ((π β πππβ‘γπ)γ)/πβcosβ‘π₯ (π₯π’π¦)β¬(π‘βπ) ((π β πππβ‘γπ)γ)/πγUsing (πππ)β¬(ββ0) π ππβ‘β/β = 1 & (πππ)β¬(ββ0) γ(1 β πππ γβ‘γβ)γ/β = 0 Using (πππ)β¬(ββ0) π ππβ‘β/β = 1 & (πππ)β¬(ββ0) γ(1 β πππ γβ‘γβ)γ/β = 0